Cycle decompositions V: Complete graphs into cycles of arbitrary lengths

Proc. London Math. Soc. Page 1 of 40 C�London Mathematical Society 2013. All rights reserved.For permissions, please e-mail: [email protected]

doi:10.1112/plms/pdt051

Cycle decompositions V: Complete graphs into cyclesof arbitrary lengths

Darryn Bryant, Daniel Horsley and William Pettersson

Abstract

We show that the complete graph on n vertices can be decomposed into t cycles of specifiedlengths m1, . . . , mt if and only if n is odd, 3 � mi � n for i = 1, . . . , t, and m1 + · · · + mt =

(n2

).

We also show that the complete graph on n vertices can be decomposed into a perfect matchingand t cycles of specified lengths m1, . . . , mt if and only if n is even, 3 � mi � n for i = 1, . . . , t,and m1 + · · · + mt =

(n2

) − n/2.

1. Introduction

A decomposition of a graph K is a set of subgraphs of K whose edge sets partition the edgeset of K. In 1981, Alspach [3] asked whether it is possible to decompose the complete graph onn vertices, denoted by Kn, into t cycles of specified lengths m1, . . . ,mt whenever the obviousnecessary conditions are satisfied; namely that n is odd, 3 � mi � n, and m1 + · · · + mt =

(n2

).

He also asked whether it is possible to decompose Kn into a perfect matching and t cycles ofspecified lengths m1, . . . ,mt whenever n is even, 3 � mi � n, and m1 + · · · + mt =

(n2

)− n/2.

Again, these conditions are obviously necessary.Here, we solve Alspach’s problem by proving the following theorem.

Theorem 1. There is a decomposition {G1, . . . , Gt} of Kn in which Gi is an mi-cycle fori = 1, . . . , t if and only if n is odd, 3 � mi � n for i = 1, . . . , t, and m1 + · · · + mt = n(n − 1)/2.There is a decomposition {G1, . . . , Gt, I} of Kn in which Gi is an mi-cycle for i = 1, . . . , t and Iis a perfect matching if and only if n is even, 3 � mi � n for i = 1, . . . , t, and m1 + · · · + mt =n(n − 2)/2.

Let K be a graph and let M = (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t.If each vertex of K has even degree, then an (M)-decomposition of K is a decomposition{G1, . . . , Gt} such that Gi is an mi-cycle for i = 1, . . . , t. If each vertex of K has odd degree,then an (M)-decomposition of K is a decomposition {G1, . . . , Gt, I} such that Gi is an mi-cyclefor i = 1, . . . , t and I is a perfect matching in K.

We say that a list (m1, . . . ,mt) of integers is n-admissible if 3 � m1, . . . ,mt � n and m1 +· · · + mt = n�(n − 1)/2�. Note that n�(n − 1)/2� =

(n2

)if n is odd, and n�(n − 1)/2� =

(n2

)−

n/2 if n is even. Thus, we can rephrase Alspach’s question as follows. Prove that for eachn-admissible list M , there exists an (M)-decomposition of Kn.

A decomposition of Kn into 3-cycles is equivalent to a Steiner triple system of order n, and adecomposition of Kn into n-cycles is a Hamilton decomposition. Thus, the work of Kirkman [37]and Walecki (see [5, 40]) from the 1800s addresses Alspach’s problem in the cases where M is

Received 6 April 2012; revised 9 August 2013.

2010 Mathematics Subject Classification 05B30, 05C70.

This research was supported by the Australian Research Council via grants DP0770400, DE120100040,DP120100790 and DP120103067. The third author was supported by an Australian Postgraduate Award.

Proceedings of the London Mathematical Society Advance Access published October 16, 2013

Page 2 of 40 DARRYN BRYANT, DANIEL HORSLEY AND WILLIAM PETTERSSON

of the form (3, 3, . . . , 3) or (n, n, . . . , n). The next results on Alspach’s problem appeared in the1960s [38, 43, 44], and a multitude of results have appeared since then. Many of these focussedon the case of decompositions into cycles of uniform length [7, 8, 12, 14, 32, 34, 35, 45], anda complete solution in this case was eventually obtained [6, 46].

There have also been many papers on the case where the lengths of the cycles in thedecomposition may vary. In recent work [18, 20, 21], the first two authors have made progressby developing methods introduced in [19, 22]. In [20], Alspach’s problem is settled in the casewhere all the cycle lengths are greater than about n/2, and in [21] the problem is completelysettled for sufficiently large odd n. Earlier results for the case of cycles of varying lengths canbe found in [1, 2, 9, 24, 25, 32, 33, 36]. See [16] for a survey on Alspach’s problem, andsee [28] for a survey on cycle decompositions generally.

The analogous problems on decompositions of complete graphs into matchings, stars or pathshave all been completely solved, see [11, 17, 39], respectively. It is also worth mentioning thatthe easier problems in which each Gi is required only to be a closed trail of length mi oreach Gi is required only to be a 2-regular graph of order mi have been solved in [10, 22, 26].Decompositions of complete multigraphs into cycles are considered in [23].

Balister [9] has verified by computer that Theorem 1 holds for n � 14, and we include thisresult as a lemma for later reference.

Lemma 2 [9]. Theorem 1 holds for n � 14.

Our proof of Theorem 1 relies heavily on the reduction of Alspach’s problem obtained in [21],see Theorem 3. Throughout the paper, we use the notation νi(M) to denote the number ofoccurrences of i in a given list M .

Definition. A list M is an n-ancestor list if it is n-admissible and satisfies

(1) ν6(M) + ν7(M) + · · · + νn−1(M) ∈ {0, 1};(2) if ν5(M) � 3, then 2ν4(M) � n − 6;(3) if ν5(M) � 2, then 3ν3(M) � n − 10;(4) if ν4(M) � 1 and ν5(M) � 1, then 3ν3(M) � n − 9;(5) if ν4(M) � 1, then νi(M) = 0 for i ∈ {n − 2, n − 1}; and(6) if ν5(M) � 1, then νi(M) = 0 for i ∈ {n − 4, n − 3, n − 2, n − 1}.

Thus, an n-ancestor list is of the form

(3, 3, . . . , 3, 4, 4, . . . , 4, 5, 5, . . . , 5, k, n, n, . . . , n),

where k is either absent or in the range 6 � k � n − 1, and there are additional constraintsinvolving the number of occurrences of cycle lengths in the list. The following theorem wasproved in [21].

Theorem 3 [21, Theorem 4.1]. For each positive integer n, if there exists an (M)-decomposition of Kn for each n-ancestor list M, then there exists an (M)-decomposition ofKn for each n-admissible list M .

Our goal is to construct an (M)-decomposition of Kn for each n-ancestor list M . We splitthis problem into two cases: the case where νn(M) � 2 and the case where νn(M) � 1. Inparticular, we prove the following two results.

CYCLE DECOMPOSITIONS V Page 3 of 40

Lemma 4. If M is an n-ancestor list with νn(M) � 2, then there is an (M)-decompositionof Kn.

Proof. See Section 3.

Lemma 5. If Theorem 1 holds for Kn−3, Kn−2 and Kn−1, then there is an (M)-decomposition of Kn for each n-ancestor list M satisfying νn(M) � 1.

Proof. The case νn(M) = 0 is proved in Section 4 (see Lemma 23) and the case νn(M) = 1is proved in Section 5 (see Lemma 45).

Lemmas 4 and 5 allow us to prove our main result using induction on n.

Proof of Theorem 1. The proof is by induction on n. By Lemma 2, Theorem 1 holds forn � 14. So, let n � 15 and assume Theorem 1 holds for complete graphs having fewer than nvertices. By Theorem 3, it suffices to prove the existence of an (M)-decomposition of Kn foreach n-ancestor list M . Lemma 4 covers each n-ancestor list M with νn(M) � 2, and using theinductive hypothesis, Lemma 5 covers those with νn(M) � 1.

2. Notation

We shall sometimes use superscripts to specify the number of occurrences of a particular integerin a list. That is, we define (mα1

1 , . . . ,mαtt ) to be the list comprised of αi occurrences of mi

for i = 1, . . . , t. Let M = (mα11 , . . . ,mαt

t ) and let M ′ = (mβ11 , . . . ,mβt

t ), where m1, . . . ,mt aredistinct. Then (M,M ′) is the list (mα1+β1

1 , . . . ,mαt+βt

t ) and, if 0 � βi � αi for i = 1, . . . , t,M − M ′ is the list (mα1−β1

1 , . . . ,mαt−βt

t ).Let Γ be a finite group and let S be a subset of Γ such that the identity of Γ is not in S and

such that the inverse of any element of S is also in S. The Cayley graph on Γ with connectionset S, denoted by Cay(Γ, S), has the elements of Γ as its vertices and there is an edge betweenvertices g and h if and only if g = hs for some s ∈ S.

A Cayley graph on a cyclic group is called a circulant graph. For any graph with vertex setZn, we define the length of an edge xy to be x − y or y − x, whichever is in {1, . . . , �n/2�}. It isconvenient to be able to describe the connection set of a circulant graph on Zn by listing onlyone of s and n − s. Thus, we use the following notation. For any subset S of Zn \ {0} such thats ∈ S and n − s ∈ S implies n = 2s, we define 〈S〉n to be the Cayley graph Cay(Zn, S ∪ −S).

Let m ∈ {3, 4, 5} and let D = {a1, . . . , am}, where a1, . . . , am are positive integers. If there isa partition {D1,D2} of D such that

∑D1 −

∑D2 = 0, then D is called a difference m-tuple.

If there is a partition {D1,D2} of D such that∑

D1 −∑

D2 = 0 (mod n), then D is called amodulo n difference m-tuple. Clearly, any difference m-tuple is also a modulo n difference m-tuple for all n. We may use the terms difference triple, quadruple and quintuple, respectively,rather than 3-tuple, 4-tuple and 5-tuple. For m ∈ {3, 4, 5}, it is clear that if D is a differencem-tuple, then there is an (mn)-decomposition of 〈D〉n for all n � 2max(D) + 1, and that if Dis a modulo n difference m-tuple, then there is an (mn)-decomposition of 〈D〉n.

We denote the complete graph with vertex set V by KV and the complete bipartite graphwith parts U and V by KU,V . If G and H are graphs then G − H is the graph with vertex setV (G) ∪ V (H) and edge set E(G) \ E(H). If G and H are graphs whose vertex sets are disjoint,then G ∨ H is the graph with vertex set V (G) ∪ V (H) and edge set E(G) ∪ E(H) ∪ {xy : x ∈V (G), y ∈ V (H)}. A cycle with m edges is called an m-cycle and is denoted (x1, . . . , xm),


where x1, . . . , xm are the vertices of the cycle and x1x2, . . . , xm−1xm, xmx1 are the edges. Apath with m edges is called an m-path and is denoted [x0, . . . , xm], where x0, . . . , xm arethe vertices of the path and x0x1, . . . , xm−1xm are the edges. A graph is said to be even ifevery vertex of the graph has even degree and is said to be odd if every vertex of the graphhas odd degree.

A packing of a graph K is a decomposition of some subgraph G of K, and the graph K − Gis called the leave of the packing. An (M)-packing of Kn is an (M)-decomposition of somesubgraph G of Kn such that G is an even graph if n is odd and G is an odd graph if n iseven (recall that an (M)-decomposition of an odd graph contains a perfect matching). Thus,the leave of an (M)-packing of Kn is an even graph and, like an (M)-decomposition of Kn, an(M)-packing of Kn contains a perfect matching if and only if n is even. A decomposition of agraph into Hamilton cycles is called a Hamilton decomposition.

3. The case of at least two Hamilton cycles

The purpose of this section is to prove Lemma 4 which states that there is an(M)-decomposition of Kn for each n-ancestor list M with νn(M) � 2. We first give a generaloutline of this proof. Theorem 1 has been proved in the case where M = (3a, nb) for somea, b � 0 (see [25]), so we will restrict our attention to ancestor lists which are not of this form.The basic construction involves decomposing Kn into 〈S〉n and Kn − 〈S〉n where, for somex � 8, the connection set S is either {1, . . . , x} or {1, . . . , x − 1} ∪ {x + 1} so that

∑S is even.

We partition any given n-ancestor list M into two lists Ms and Ms = M − Ms, and construct an(Ms)-decomposition of 〈S〉n and an (Ms)-decomposition of Kn − 〈S〉n. This yields the desired(M)-decomposition of Kn. Taking S = {1, . . . , x − 1} ∪ {x + 1}, rather than S = {1, . . . , x}, isnecessary when 1 + · · · + x is odd as many desired cycle decompositions of 〈{1, . . . , x}〉n donot exist when 1 + · · · + x is odd, see [27].

If M = (3α3n+β3 , 4α4n+β4 , 5α5n+β5 , kγ , nδ), where αi � 0 and 0 � βi � n − 1 for i ∈ {3, 4, 5},6 � k � n − 1, γ ∈ {0, 1}, and δ � 2, then we usually choose Ms = (3β3 , 4β4 , 5β5 , kγ). However,if this would result in

∑Ms being less than 4n, then we sometimes adjust this definition slightly.

We always choose Ms such that∑

Ms is at most 8n, which explains why we have |S| � 8.Our (Ms)-decompositions of 〈S〉n will be constructed using adaptations of techniques used

in [27, 29]. We construct our (Ms)-decompositions of Kn − 〈S〉n using a combination ofdifference methods and results on Hamilton decompositions of circulant graphs. In general,we split the problem into the case ν5(M) � 2 and the case ν5(M) � 3. In the former case itwill follow from our choice of Ms that Ms = (3tn, 4qn, nh) for some t, q, h � 0 and in the lattercase it will follow from our choice of Ms that Ms = (5rn, nh) for some r, h � 0.

The precise definition of Ms is given in Lemma 6, which details the properties that we requireof our partition of M into Ms and Ms, and establishes its existence. The definition includesseveral minor technicalities in order to deal with complications and exceptions that arise in theabove-described approach. Throughout the remainder of this section, for a given n-ancestorlist M such that νn(M) � 2 and M = (3a, nb) for any a, b � 0, we shall use the notation Ms

and Ms to denote the lists constructed in the proof of Lemma 6. If νn(M) � 1 or M = (3a, nb)for some a, b � 0, then Ms and Ms are not defined.

Lemma 6. If M is any n-ancestor list such that νn(M) � 2 and M = (3a, nb) for anya, b � 0, then there exists a partition of M into sublists Ms and Ms such that

(1)∑

Ms ∈ {2n, 3n, . . . , 8n} and∑

Ms = 8n when ν5(M) � 2;(2) if

∑Ms = 2n, then νn(Ms) = 1 and Ms = (nh) for some h � 1;

(3) if∑

Ms = 3n, then νn(Ms) ∈ {0, 1} and Ms = (nh) for some h � 1;


(4) if∑

Ms ∈ {4n, 5n, . . . , 8n} and ν5(M) � 3, then νn(Ms) = 0 and Ms = (5rn, nh) forsome r � 0, h � 2;

(5) if∑

Ms ∈ {4n, 5n, . . . , 7n} and ν5(M) � 2, then νn(Ms) = 0 and Ms = (3tn, 4qn, nh) forsome t, q � 0, h � 2; and

(6) Ms = (35n/3).

Proof. Let M be an n-ancestor list. The conditions of the lemma imply n � 7. We will firstdefine a list Me which in many cases will serve as Ms, but will sometimes need to be adjustedslightly.

If

M = (3α3n+β3 , 4α4n+β4 , 5α5n+β5 , kγ , nδ),

where αi � 0 and 0 � βi � n − 1 for i ∈ {3, 4, 5}, 6 � k � n − 1, γ ∈ {0, 1}, and δ � 2, then

Me = (3β3 , 4β4 , 5β5 , kγ).

It is clear from the definition of n-ancestor list that if we take Ms = Me, then (4) and (5) aresatisfied.

We now show that∑

Me ∈ {0, n, 2n, . . . , 8n}, and that∑

Me = 8n when ν5(M) � 2. Notingthat

∑Me � 3β3 + 4β4 + 5β5 + (n − 1) and separately considering the cases ν5(M) � 3,

ν5(M) ∈ {1, 2} and ν5(M) = 0, it is routine to use the definition of (M)-ancestor lists toshow that

∑Me < 9n, and that

∑Me < 8n when ν5(M) � 2. Thus, because it follows

from∑

M = n�(n − 1)/2� and the definition of Me that n divides∑

Me, we have that∑Me ∈ {0, n, 2n, . . . , 8n}, and that

∑Me = 8n when ν5(M) � 2.

If∑

Me ∈ {4n, 5n, 6n, 7n, 8n}, then we let Ms = Me. If∑

Me ∈ {0, n, 2n, 3n}, then we defineMs by

Ms =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(Me, 4n) if α4 > 0;(Me, 5n) if α4 = 0 and α5 > 0;(Me, 3n) if α4 = α5 = 0 and α3 > 0;(Me, n) if α3 = α4 = α5 = 0 and

∑Me ∈ {n, 2n};

Me otherwise.

Using the definition of Ms and the fact that M is an n-ancestor list with M = (3a, nb) for anya, b � 0, it is routine to check that Ms satisfies (1)–(6).

Before proving Lemma 4, we need a number of preliminary lemmas. The first three give usthe necessary decompositions of 〈S〉n where S = {1, . . . , x} or S = {1, . . . , x − 1} ∪ {x + 1} forsome x � 8. Lemma 8 was proved independently in [15, 42], and is a special case of Theorem 5in [27]. Lemmas 7 and 9 will be proved in Section 6.

Lemma 7. If

S ∈ {{1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7},{1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7, 8}},

n � 2max(S) + 1, and M = (m1, . . . ,mt, k) is any list satisfying mi ∈ {3, 4, 5} for i = 1, . . . , t,3 � k � n, and

∑M = |S|n, then there is an (M)-decomposition of 〈S〉n, except possibly

when

(i) S = {1, 2, 3, 4, 6}, n ≡ 3 (mod 6) and M = (35n/3); or

(ii) S = {1, 2, 3, 4, 6}, n ≡ 4 (mod 6) and M = (3(5n−5)/3, 5).



Lemma 8 [15, 42]. If n � 5 and M = (m1, . . . ,mt, n) is any list satisfying mi ∈ {3, . . . , n}for i = 1, . . . , t, and

∑M = 2n, then there is an (M)-decomposition of 〈{1, 2}〉n.

Lemma 9. If n � 7 and M = (m1, . . . ,mt, k, n) is any list satisfying mi ∈ {3, 4, 5} for i =1, . . . , t, 3 � k � n, and

∑M = 3n, then there is an (M)-decomposition of 〈{1, 2, 3}〉n.


We now present the lemmas which give us the necessary decompositions of Kn − 〈S〉n.Lemma 10 was proved in [25] where it was used to prove Theorem 1 in the case where M =(3a, nb) for some a, b � 0. Lemmas 11 and 12 give our main results on decompositions ofKn − 〈S〉n. Lemma 11 is for the case ν5(M) � 2 and Lemma 12 is for the case ν5(M) � 3.

Lemma 10 [25, Lemma 3.1]. If 1 � h � �(n − 1)/2�, then there is an (nh)-decompositionof Kn − 〈{1, . . . , �(n − 1)/2� − h}〉n.

Lemma 11. If S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}} and n �2max(S) + 1, t � 0, q � 0 and h � 2 are integers satisfying 3t + 4q + h = �(n − 1)/2� − |S|,then there is a (3tn, 4qn, nh)-decomposition of Kn − 〈S〉n, except possibly when h = 2, S ={1, 2, 3, 4, 5, 6, 7} and

(i) n ∈ {25, 26} and t = 1; or(ii) n = 31 and t = 2.


Lemma 12. If S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5,6, 7, 8}} and n � 2max(S) + 1, r � 0 and h � 2 are integers satisfying 5r + h = �(n − 1)/2� −|S|, then there is a (5rn, nh)-decomposition of Kn − 〈S〉n.


We also need Lemmas 14 and 15 to deal with cases arising from the possible exceptionsin Lemmas 7 and 11, respectively. To prove Lemma 14 we use the following special case ofLemma 2.8 in [21].

Lemma 13. If there exists an (M, 42)-decomposition of Kn in which there are two 4-cyclesintersecting in exactly one vertex, then there exists an (M, 3, 5)-decomposition of Kn.

Lemma 14. If M is an n-ancestor list such that νn(M) � 2, Ms = (3(5n−5)/3, 5) andn ≡ 4 (mod 6), then there is an (M)-decomposition of Kn.


Proof. We will construct an (Ms, 3(5n−8)/3, 42)-decomposition of Kn in which two 4-cyclesintersect in exactly one vertex. The required (M)-decomposition of Kn can then be obtainedby applying Lemma 13.

By Lemma 11, there is an (Ms)-decomposition of Kn − 〈{1, 2, 3, 4, 6}〉n, so it sufficesto construct a (3(5n−8)/3, 42)-decomposition of 〈{1, 2, 3, 4, 6}〉n in which the two 4-cyclesintersect in exactly one vertex for all n ≡ 4 (mod 6) with n � 16 (note that the conditionsof the lemma imply n � 16). The union of the following two sets of cycles gives such adecomposition.

{(0, 4, 2, 6), (2, 3, 5, 8), (1, 5, 7), (3, 4, 7), (3, 6, 9), (4, 5, 6)}

{(x + 6i, y + 6i, z + 6i) : i ∈{

0, . . . ,(n − 10)

6

},

(x, y, z) ∈ {(4, 8, 10), (5, 9, 11), (6, 8, 12), (6, 7, 10), (7, 11, 13), (7, 8, 9),(9, 12, 15), (9, 10, 13), (10, 11, 12), (8, 11, 14)}}.

Lemma 15. If M is an n-ancestor list such that ν5(M) � 2, νn(M) = 2,∑

Ms = 7n, and

(i) n = 25 and ν3(Ms) = 25;(ii) n = 26 and ν3(Ms) = 26; or(iii) n = 31 and ν3(Ms) = 62;

then there is an (M)-decomposition of Kn.

Proof. We begin by showing that it is possible to partition Ms into two lists M1s and

M2s such that

∑M1

s = 3n and∑

M2s = 4n. If ν3(Ms) � n or ν4(Ms) � n, then clearly such a

partition exists. Otherwise, νn(Ms) = 0 by Property (5) of Lemma 6, and so by the definitionof n-ancestor list and the hypotheses of this lemma, we have that

7n =∑

Ms � 3ν3(Ms) + 4ν4(Ms) + 10 + (n − 1).

It is routine to check, using 3ν3(Ms) � 3n − 3 and 4ν4(Ms) � 4n − 4, that ν4(Ms) � (3n − 6)/4and ν3(Ms) � (2n − 5)/3. Thus, for n = 25, n = 26 and n = 31, we can choose M1

s = (3, 418),M1

s = (32, 418), and M1s = (33, 421), respectively. This yields the desired partition of Ms.

For n = 25 we note that 〈{1, 2, 3}〉n ∼= 〈{2, 4, 6}〉n (with x �→ 2x being an isomorphism)and 〈{1, 2, 3, 4}〉n ∼= 〈{1, 7, 8, 9}〉n (with x �→ 8x being an isomorphism). Since {3, 10, 12} isa modulo 25 difference triple and 〈{5, 11}〉25 has a Hamilton decomposition (by a result ofBermond et al. [13], see Lemma 53), this gives us a decomposition of K25 into a copyof 〈{1, 2, 3}〉25, a copy of 〈{1, 2, 3, 4}〉25, twenty-five 3-cycles and two Hamilton cycles. ByLemma 7, there is an (M1

s )-decomposition of 〈{1, 2, 3}〉25 and an (M2s )-decomposition of

〈{1, 2, 3, 4}〉25, and this gives us the required (M)-decomposition of K25.For n = 26, we note that 〈{1, 2, 3, 4}〉n ∼= 〈{5, 6, 10, 11}〉n (with x �→ 5x being an isomor-

phism). Since {4, 8, 12} is a difference triple and 〈{7, 9}〉26 has a Hamilton decomposition (bya result of Bermond et al. [13], see Lemma 53), this gives us a decomposition of K26 intoa copy of 〈{1, 2, 3}〉26, a copy of 〈{1, 2, 3, 4}〉26, twenty-six 3-cycles and two Hamilton cycles.By Lemma 7, there is an (M1

s )-decomposition of 〈{1, 2, 3}〉26 and an (M2s )-decomposition of

〈{1, 2, 3, 4}〉26, and this gives us the required (M)-decomposition of K26.For n = 31, we note that 〈{1, 2, 3, 4}〉n ∼= 〈{4, 8, 12, 15}〉n (with x �→ 4x being an isomor-

phism). Since {5, 6, 11} is a difference triple, {7, 10, 14} is a modulo 31 difference triple,


and 〈{9, 13}〉31 has a Hamilton decomposition (by a result of Bermond et al. [13], seeLemma 53), this gives us a decomposition of K31 into a copy of 〈{1, 2, 3}〉31, a copyof 〈{1, 2, 3, 4}〉31, sixty-two 3-cycles and two Hamilton cycles. By Lemma 7, there is an(M1

s )-decomposition of 〈{1, 2, 3}〉31 and an (M2s )-decomposition of 〈{1, 2, 3, 4}〉31, which yields

required (M)-decomposition of K31.

We can now prove Lemma 4 which states that if M is an n-ancestor list with νn(M) � 2,then there is an (M)-decomposition of Kn.

Proof of Lemma 4. If M = (3a, nb) for some a, b � 0, then we can use the main resultfrom [25] to obtain an (M)-decomposition of Kn, so we can assume that M = (3a, nb) for anya, b � 0. By Lemma 2, we can assume that n � 15. Partition M into Ms and Ms. The proofsplits into cases according to the value of

∑Ms, which by Lemma 6 is in {2n, 3n, . . . , 8n}.

Case 1. Suppose that∑

Ms = 2n. In this case, from Property (2) of Lemma 6 we haveνn(Ms) = 1 and Ms = (nh) for some h � 1. The required decomposition of Kn can be obtainedby combining an (Ms)-decomposition of 〈{1, 2}〉n (which exists by Lemma 8) with a Hamiltondecomposition of Kn − 〈{1, 2}〉n (which exists by Lemma 10).


Ms = 3n. In this case, from Property (3) of Lemma 6 we haveνn(Ms) ∈ {0, 1} and Ms = (nh) for some h � 1. The required decomposition of Kn can beobtained by combining an (Ms)-decomposition of 〈{1, 2, 3}〉n (which exists by Lemma 7 or 9)with a Hamilton decomposition of Kn − 〈{1, 2, 3}〉n (which exists by Lemma 10).


Ms ∈ {4n, 5n, 6n, 7n, 8n} and ν5(M) � 3. In this case, from Property(4) of Lemma 6, we have νn(Ms) = 0 and Ms = (5rn, nh) for some r � 0, h � 2, and we alsohave 3ν3(M) � n − 10 from the definition of n-ancestor list. We let

S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7, 8}}

such that |S| = (1/n)∑

Ms and obtain the required decomposition of Kn by combining an(Ms)-decomposition of 〈S〉n (which exists by Lemma 7), with an (Ms)-decomposition of Kn −〈S〉n (which exists by Lemma 12). Note that the condition 3ν3(M) � n − 10 implies that therequired (Ms)-decomposition of 〈S〉n is not among the listed possible exceptions in Lemma 7.Note also that the condition n � 2max(S) + 1 required in Lemmas 7 and 12 is easily seen tobe satisfied because n � 15 and

∑Ms � n�(n − 1)/2�.


Ms ∈ {4n, 5n, 6n, 7n, 8n} and ν5(M) � 2. In this case, we haveνn(Ms) = 0 and Ms = (3tn, 4qn, nh) for some t, q � 0, h � 2 (see Property (5) in Lemma 6),and

∑Ms = 8n (see Property (1) in Lemma 6). We let

S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}}

such that |S| = (1/n)∑

Ms. If Lemma 7 gives us an (Ms)-decomposition of 〈S〉n and Lemma 11gives us an (Ms)-decomposition of Kn − 〈S〉n, then we have the required decomposition of Kn.The condition n � 2max(S) + 1 required in Lemmas 7 and 11 is satisfied because n � 15. Thisleaves only the cases arising from the possible exceptions in Lemmas 7 and 11, and these arecovered by Lemmas 14 and 15, respectively.

4. The case of no Hamilton cycles

In this section we prove that Lemma 5 holds in the case νn(M) = 0. In this case, for n � 15, oneof ν3(M), ν4(M) and ν5(M) must be sizable, and the proof splits into three cases accordingly.Each of these three cases splits into subcases according to whether n is even or odd. In each casewe construct the required decomposition of Kn from a suitable decomposition of Kn−1 or Kn−2.


4.1. Many 3-cycles and no Hamilton cycles

In Lemma 16 we construct the required decompositions of complete graphs of odd order, andin Lemma 17 we construct the required decompositions of complete graphs of even order.

Lemma 16. If n is odd, Theorem 1 holds for Kn−1, and (M, 3(n−1)/2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 3(n−1)/2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since (M, 3(n−1)/2) is an n-ancestor listwith νn(M) = 0, it follows that M is (n − 1)-admissible. Thus, by assumption there is an(M)-decomposition D of KU . Let I be the perfect matching in D. Then

D ∪D1

is an (M, 3(n−1)/2)-decomposition of KV , where D1 is a (3(n−1)/2)-decomposition ofK{∞} ∨ I.

Lemma 17. If n is even, Theorem 1 holds for Kn−1, and (M, 3(n−2)/2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 3(n−2)/2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since (M, 3(n−2)/2) is an n-ancestor listwith νn(M) = 0, it follows that (M,n − 2) is (n − 1)-admissible and so by assumption thereis an (M,n − 2)-decomposition D of KU . Let C be an (n − 2)-cycle in D, let {I, I1} be adecomposition of C into two matchings, and let x be the vertex in U \ V (C). Then

(D \ {C}) ∪ {I + ∞x} ∪ D1

is an (M, 3(n−2)/2)-decomposition of KV , where D1 is a (3(n−2)/2)-decomposition ofK{∞} ∨ I1.


Lemma 18. If n is odd, Theorem 1 holds for Kn−2, and (M, 4(n+1)/2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 4(n+1)/2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 2,let ∞1 and ∞2 be distinct vertices not in U , and let V = U ∪ {∞1,∞2}. Since (M, 4(n+1)/2)is an n-ancestor list with νn(M) = 0, it follows from (5) in the definition of ancestor lists thatany cycle length in M is at most n − 3. Thus, it is easily seen that (M, 5) is (n − 2)-admissibleand by assumption there is an (M, 5)-decomposition D of KU .

Let C be a 5-cycle in D and let x, y and z be vertices of C such that x and y are adjacentin C and z is not adjacent to either x or y in C. Then

(D \ {C}) ∪ D1 ∪ D2

is an (M, 4(n+1)/2)-decomposition of KV , where

(i) D1 is a (4(n−5)/2)-decomposition of K{∞1,∞2},U\{x,y,z}; and(ii) D2 is a (43)-decomposition of K{∞1,∞2},{x,y,z} ∪ [∞1,∞2] ∪ C.

These decompositions are straightforward to construct.


Lemma 19. If n is even, Theorem 1 holds for Kn−2, and (M, 4(n−2)/2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 4(n−2)/2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n − 2,let ∞1 and ∞2 be distinct vertices not in U , and let V = U ∪ {∞1,∞2}. Since (M, 4(n−2)/2)is an n-ancestor list with νn(M) = 0, it follows from (5) in the definition of ancestor lists thatany cycle length in M is at most n − 3. Thus, it is easily seen that M is (n − 2)-admissible andby assumption there is an (M)-decomposition D of KU . Let I be the perfect matching in D.Then

(D \ {I}) ∪ {I + ∞1∞2} ∪ D1

is an (M, 4(n−2)/2)-decomposition of KV , where D1 is a (4(n−2)/2)-decomposition ofK{∞1,∞2},U .


We will make use of the following lemma in this subsection and in Subsection 5.5.

Lemma 20. If G is a 3-regular graph which contains a perfect matching and ∞ is a vertexnot in V (G), then there is a decomposition of K{∞} ∨ G into 1

2 |V (G)| 5-cycles.

Proof. Let I be a perfect matching in G. Then G − I is a 2-regular graph on the vertex setV (G) and hence it can be given a coherent orientation O. Let

D = {(∞, a, b, c, d) : bc ∈ E(I) and (b, a), (c, d) ∈ E(O)}be a set of (undirected) 5-cycles. Because O contains exactly one arc directed from each vertex ofV (G), |D| = |E(I)| = 1

2 |V (G)| and each edge of G appears in exactly one cycle in D. Further,because O contains exactly one arc directed to each vertex of V (G), each edge of K{∞},V

appears in exactly one cycle in D. Thus, D is a decomposition of K{∞} ∨ G into 12 |V (G)|

5-cycles.

Lemma 21. If n is odd, Theorem 1 holds for Kn−1, and (M, 5(n−1)/2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 5(n−1)/2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since the list (M,n − 1) is easily seen tobe (n − 1)-admissible, by assumption there is an (M,n − 1)-decomposition D of KU . Let C bean (n − 1)-cycle in D and let I be the perfect matching in D. Then

(D \ {C, I}) ∪ D1

is an (M, 5(n−1)/2)-decomposition of KV , where D1 is a (5(n−1)/2)-decomposition of K{∞} ∨(C ∪ I) (this exists by Lemma 20).

Lemma 22. If n is even, Theorem 1 holds for Kn−2, and (M, 5n−2) is an n-ancestor listwith νn(M) = 0, then there is an (M, 5n−2)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n − 2,let ∞1 and ∞2 be distinct vertices not in U , and let V = U ∪ {∞1,∞2}. Since (M, 5n−2) is


an n-ancestor list with νn(M) = 0, it follows from (6) in the definition of ancestor lists thatany cycle length in M is at most n − 5. Thus, it is easily seen that the list (M, (n − 2)3) is(n − 2)-admissible and by assumption there is an (M, (n − 2)3)-decomposition D of KU . LetC1, C2 and C3 be distinct (n − 2)-cycles in D and let I be the perfect matching in D. Let{I1, I2} be a decomposition of C3 into two perfect matchings. Then

(D \ {C1, C2, C3, I}) ∪ {I + ∞1∞2} ∪ D1 ∪ D2

is an (M, 5n−2)-decomposition of KV , where for i = 1, 2, Di is a (5(n−2)/2)-decomposition ofK{∞i} ∨ (Ci ∪ Ii) (these exist by Lemma 20).

4.4. Proof of Lemma 5 in the case of no Hamilton cycles

Lemma 23. If Theorem 1 holds for Kn−1 and Kn−2, then there is an (M)-decompositionof Kn for each n-ancestor list M satisfying νn(M) = 0.

Proof. By Lemma 2, we can assume that n � 15. If there is a cycle length in M which isat least 6 and at most n − 1, then let k be this cycle length. Otherwise let k = 0. We dealseparately with the case n is odd and the case n is even.

Case 1. Suppose that n is odd. Since n � 15 and 3ν3(M) + 4ν4(M) + 5ν5(M) + k =n(n − 1)/2, it can be seen that either ν3(M) � (n − 1)/2, ν4(M) � (n + 1)/2 or ν5(M) �(n − 1)/2. If ν3(M) � (n − 1)/2, then the result follows by Lemma 16. If ν4(M) � (n + 1)/2,then the result follows by Lemma 18. If ν5(M) � (n − 1)/2, then the result follows byLemma 21.

Case 2. Suppose that n is even. Since n � 16, 3ν3(M) + 4ν4(M) + 5ν5(M) + k = n(n − 2)/2and k � n − 1, it can be seen that either ν3(M) � (n − 2)/2, ν4(M) � (n − 2)/2 or ν5(M) �n − 2. (To see this consider the cases ν5(M) � 3 and ν5(M) � 2 separately and use thedefinition of n-ancestor list.) If ν3(M) � (n − 2)/2, then the result follows by Lemma 17. Ifν4(M) � (n − 2)/2, then the result follows by Lemma 19. If ν5(M) � n − 2, then the resultfollows by Lemma 22.

5. The case of exactly one Hamilton cycle

In this section we prove that Lemma 5 holds in the case νn(M) = 1. Again in this case, forn � 15, one of ν3(M), ν4(M) and ν5(M) must be sizable, and the proof splits into casesaccordingly. The case in which ν3(M) is sizable further splits according to whether ν4(M) � 1,ν5(M) � 1, or ν4(M) = ν5(M) = 0. We first require some preliminary definitions and results.

5.1. Preliminaries

Let P be an (M)-packing of Kn, let P ′ be an (M ′)-packing of Kn and let S be a subset ofV (Kn). We say that P and P ′ are equivalent on S if we can write {G ∈ P : V (G) ∩ S = ∅} ={G1, . . . , Gt} and {G ∈ P ′ : V (G) ∩ S = ∅} = {G′

1, . . . , G′t} such that

(i) for i ∈ {1, . . . , t}, Gi is isomorphic to G′i;

(ii) for each x ∈ S and for i ∈ {1, . . . , t}, x ∈ V (Gi) if and only if x ∈ V (G′i); and

(iii) for all distinct x, y ∈ S and for i ∈ {1, . . . , t}, xy ∈ E(Gi) if and only if xy ∈ E(G′i).

The following lemma is from [20]. It encapsulates a key edge swapping technique which wasused in many of the proofs in [21], and which we shall make use of in this section.


Lemma 24 [20, Lemma 2.1]. Let n be a positive integer, let M be a list of integers,let P be an (M)-packing of Kn with a leave, L say, let α and β be vertices of L, letπ be the transposition (αβ), and let Z = Z(P, α, β) = (NbdL(α) ∪ NbdL(β)) \ ((NbdL(α) ∩NbdL(β)) ∪ {α, β}). Then there exists a partition of the set Z into pairs such that for eachpair {u, v} of the partition, there exists an (M)-packing of Kn, P ′ say, with a leave, L′ say,which differs from L only in that αu, αv, βu and βv are edges in L′ if and only if they arenot edges in L. Furthermore, if P = {C1, . . . , Ct} (n odd) or P = {I, C1, . . . , Ct} (n even)where C1, . . . , Ct are cycles and I is a perfect matching, then P ′ = {C ′

1, . . . , C′t} (n odd) or

P ′ = {I ′, C ′1, . . . , C

′t} (n even) where for i = 1, . . . , t, C ′

i is a cycle of the same length as Ci andI ′ is a perfect matching such that

(i) either I ′ = I or I ′ = π(I);(ii) for i = 1, . . . , t if neither α nor β is in V (Ci), then C ′

i = Ci;(iii) for i = 1, . . . , t if exactly one of α and β is in V (Ci), then either C ′

i = Ci or C ′i = π(Ci);

and(iv) for i = 1, . . . , t if both α and β are in V (Ci), then C ′

i ∈ {Ci, π(Ci), π(Pi) ∪ P †i , Pi ∪

π(P †i )} where Pi and P †

i are the two paths in Ci which have endpoints α and β.

We say that P ′ is the (M)-packing obtained from P by performing the (α, β)-switch withorigin u and terminus v (we could equally call v the origin and u the terminus). For our purposeshere, it is important to note that P ′ is equivalent to P on V (L) \ {α, β}.

We will also make use of three lemmas from [21]. The original version of Lemma 25(Lemma 2.15 in [21]) does not include the claim that P ′ is equivalent to P on V (L) \ {a, b},this follows directly from the fact that the proof uses only (a, b)-switches.

Lemma 25. Let n be a positive integer and let M be a list of integers. Suppose that thereexists an (M)-packing P of Kn with a leave L which contains two vertices a and b such thatdegL(a) + 2 � degL(b). Then there exists an (M)-packing P ′ of Kn, which is equivalent toP on V (L) \ {a, b}, and which has a leave L′ such that degL′(a) = degL(a) + 2, degL′(b) =degL(b) − 2 and degL′(x) = degL(x) for all x ∈ V (L) \ {a, b}. Furthermore,

(i) if a and b are adjacent in L, then L′ has the same number of non-trivial componentsas L;

(ii) if degL(a) = 0 and b is not a cut-vertex of L, then L′ has the same number of non-trivialcomponents as L; and

(iii) if degL(a) = 0, then either L′ has the same number of non-trivial components as L, orL′ has one more non-trivial component than L.

Similarly, the original versions of Lemmas 26 and 27 (Lemmas 2.14 and 2.11, respectively,in [21]) did not include the claims that the final decompositions are equivalent to the initialpackings on V \ U . However, these claims can be seen to hold as the proofs of the lemmasgiven in [21] require switching only on vertices of positive degree in the leave, with oneexception which we discuss shortly. The lemmas below each contain the additional hypothesisthat degL(x) = 0 for all x ∈ V \ U , and this ensures all the switches are on vertices of U andhence that the final decomposition is equivalent to the initial packing on V \ U .

The exception mentioned above occurs in the proof of the original version of Lemma 27 wherea switch on a vertex of degree 0 in the leave is required when 3 ∈ {m1,m2}. We can ensurethis switch is on a vertex in U because we have the additional hypothesis that degL(x) = 0for some x ∈ U when 3 ∈ {m1,m2}. This additional hypothesis also allows us to omit thehypothesis, included in the original version of Lemma 27, that the size of the leave be at mostn + 1, because in the proof this was used only to ensure the existence of a vertex of degree 0


in the leave when 3 ∈ {m1,m2}. Thus the modified versions stated below hold by the proofspresented in [21].

Lemma 26. Let V be a vertex set and let U be a subset of V . Let M be a list of integers andlet k, m1 and m2 be positive integers such that m1,m2 � max({3, k + 1}). Suppose that thereexists an (M)-packing P of KV with a leave L of size m1 + m2 such that Δ(L) = 4, exactly onevertex of L has degree 4, L has exactly k non-trivial components, L does not have a decom-position into two odd cycles if m1 and m2 are both even, and degL(x) = 0 for all x ∈ V \ U .Then there exists an (M,m1,m2)-decomposition of KV which is equivalent to P on V \ U .

Lemma 27. Let V be a vertex set and let U be a subset of V . Let M be a list of integers andlet m1 and m2 be integers such that m1,m2 � 3. Suppose that there exists (M)-packing P ofKV with a leave L of size m1 + m2 such that Δ(L) = 4, exactly two vertices of L have degree 4,L has exactly one non-trivial component, degL(x) = 0 for all x ∈ V \ U, and degL(x) = 0 forsome x ∈ U if 3 ∈ {m1,m2}. Then there exists an (M,m1,m2)-decomposition of KV which isequivalent to P on V \ U .

We also require Lemma 28, which deals with some small order cases.

Lemma 28. Let n be an integer such that n ∈ {15, 16, 17, 18, 19, 20, 22, 24, 26} and let Mbe an n-ancestor list such that νn(M) = 1, ν5(M) � 3, and ν4(M) � 2 if n = 24. Then thereis an (M)-decomposition of Kn.

Proof. If there is a cycle length in M which is at least 6 and at most n − 1, then let kbe this cycle length. Otherwise, let k = 0. Note that 3ν3(M) + 4ν4(M) + 5ν5(M) + k + n =n�(n − 1)/2� and that, because M is an n-ancestor list with ν5(M) � 3, it follows that3ν3(M) � n − 10, 2ν4(M) � n − 6, and k � n − 5.

Using this, it is routine to check that if n = 15, then M must be one of 12 possible lists andif n = 16, then M must be one of 26 possible lists. In each of these cases, we have constructedan (M)-decomposition of Kn by computer search.

If n ∈ {17, 18, 19, 20, 22, 24, 26}, then we partition {1, . . . , �n/2�} into sets S1, S2 and S3

according to the following table:

n S1 S2 S3

17 {1, 2, 3, 4, 5, 6, 7} ∅ {8}18 {1, 2, 3, 4, 5, 6, 7} ∅ {8, 9}19 {1, 2, 3, 4, 5, 6, 7, 8} ∅ {9}20 {1, 2, 3, 4, 5, 6, 7, 8} ∅ {9, 10}22 {1, 2, 3, 4} {6, 7, 8, 9, 10} {5, 11}24 {1, 2, 3} {4, 6, 7, 8, 11} {5, 9, 10, 12}26 {1, 2, 3, 4, 5, 7} {6, 8, 9, 10, 11} {12, 13}

Using 3ν3(M) � n − 10, 2ν4(M) � n − 6 and k � n − 5, it is routine to check that ν5(M) �n when n ∈ {22, 26}, and that ν5(M) � n + 8 when n = 24. By Lemma 7, there is an (M ′)-decomposition of 〈S1〉n, where M = (M ′, n) when n ∈ {17, 18, 19, 20}, M = (M ′, 5n, n) whenn ∈ {22, 26}, and M = (M ′, 42, 5n+8, n) when n = 24. For n ∈ {22, 24, 26}, it is easy to seethat S2 is a modulo n difference 5-tuple, and so there is a (5n)-decomposition of 〈S2〉n. Forn ∈ {17, 18, 19, 20, 22, 26}, there is an (n)-decomposition of the graph 〈S3〉n, as it is either ann-cycle or a connected 3-regular Cayley graph on a cyclic group, and the latter are well knownto contain a Hamilton cycle, see [30]. For n = 24, 〈{5, 9, 10}〉n ∼= 〈{1, 2, 3}〉n (with x �→ 5x being


an isomorphism). Thus, by Lemma 9 there is a (42, 58, 24)-decomposition of 〈{5, 9, 10, 12}〉24(as 〈{12}〉24 is a perfect matching). Combining these decompositions of 〈S1〉n, 〈S2〉n and 〈S3〉ngives us the required (M)-decomposition of Kn.

5.2. Many 3-cycles, one Hamilton cycle, and at least one 4- or 5-cycle

In Lemmas 29 and 30 we construct the required decompositions of complete graphs of oddorder in the cases where the decomposition contains at least one 4-cycle or at least one 5-cycle,respectively. In Lemmas 31 and 32 we construct the required decompositions of complete graphsof even order in the cases where the decomposition contains at least one 4-cycle or at least one5-cycle, respectively. These results are proved by constructing the required decomposition ofKn from a suitable decomposition of Kn−1, Kn−2 or Kn−3.

Lemma 29. If n is odd, Theorem 1 holds for Kn−1, and (M, 3(n−5)/2, 4, n) is an n-ancestorlist with νn(M) = 0, then there is an (M, 3(n−5)/2, 4, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since (M, 3(n−5)/2, 4, n) is an n-ancestorlist with νn(M) = 0, it follows that (M,n − 2) is (n − 1)-admissible and by assumption thereis an (M,n − 2)-decomposition D of KU .

Let H be an (n − 2)-cycle in D, let I be the perfect matching in D, and let [w, x, y, z] be apath in I ∪ H such that w /∈ V (H), wx, yz ∈ E(I) and xy ∈ E(H). Then

(D \ {I,H}) ∪ {H ′, (∞, x, y, z)} ∪ D1

is an (M, 3(n−5)/2, 4, n)-decomposition of KV , where

(i) H ′ = (H − [x, y]) ∪ [x,w,∞, y]; and(ii) D1 is a (3(n−5)/2)-decomposition of K{∞},U\{w,x,y,z} ∪ (I − {wx, yz}).

Lemma 30. If n is odd, Theorem 1 holds for Kn−1, and (M, 3(n−5)/2, 5, n) is an n-ancestorlist with νn(M) = 0, then there is an (M, 3(n−5)/2, 5, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since (M, 3(n−5)/2, 5, n) is an n-ancestor listwith νn(M) = 0, it follows that (M,n − 1) is (n − 1)-admissible and so by assumption there isan (M,n − 1)-decomposition D of KU .

Let H be an (n − 1)-cycle in D, let I be the perfect matching in D, and let [w, x, y, z] be apath in I ∪ H such that wx, yz ∈ E(I) and xy ∈ E(H). Then

(D \ {I,H}) ∪ {H ′, (∞, w, x, y, z)} ∪ D1


(i) H ′ = (H − [x, y]) ∪ [x,∞, y]; and(ii) D1 is a (3(n−5)/2)-decomposition of K{∞},U\{w,x,y,z} ∪ (I − {wx, yz}).

Lemma 31. If n is even, Theorem 1 holds for Kn−3, and (M, 3(3n−14)/2, 4, n) is an n-ancestor list with νn(M) = 0, then there is an (M, 3(3n−14)/2, 4, n)-decomposition of Kn.


Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n −3, let ∞1, ∞2 and ∞3 be distinct vertices not in U , and let V = U ∪ {∞1,∞2,∞3}. Since(M, 3(3n−14)/2, 4, n) is an n-ancestor list with νn(M) = 0, it follows from (5) in the definitionof ancestor lists that any cycle length in M is at most n − 3. Thus, it is easily seen that(M, (n − 4)2, n − 3) is (n − 3)-admissible and so by assumption there is an (M, (n − 4)2, n −3)-decomposition D of KU .

Let C1 and C2 be distinct (n − 4)-cycles in D, let H be an (n − 3)-cycle in D, let {I, I1}be a decomposition of C1 into two matchings, let {I2, I3} be a decomposition of C2 into twomatchings, let w be the vertex in U \ V (C1), and let [x, y, z] be a path in H ∪ I3 such thatx /∈ V (C2), xy ∈ E(H) and yz ∈ E(I3) (possibly w ∈ {x, y, z}). Then

(D \ {H,C1, C2}) ∪ {I + {∞1w,∞2∞3},H ′, (∞3, x, y, z)} ∪ D1 ∪ D2 ∪ D3

is an (M, 3(3n−14)/2, 4, n)-decomposition of KV , where

(i) H ′ = (H − [x, y]) ∪ [x,∞2,∞1,∞3, y];(ii) for i = 1, 2, Di is a (3(n−4)/2)-decomposition of K{∞i} ∨ Ii; and(iii) D3 is a (3(n−6)/2)-decomposition of K{∞3},U\{x,y,z} ∪ (I3 − yz).

Lemma 32. If n is even, Theorem 1 holds for Kn−1, and (M, 3(n−6)/2, 5, n) is an n-ancestorlist with νn(M) = 0, then there is an (M, 3(n−6)/2, 5, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n − 1,let ∞ be a vertex not in U , and let V = U ∪ {∞}. Since (M, 3(n−6)/2, 5, n) is an n-ancestor listwith νn(M) = 0, it follows that (M,n − 2, n − 1) is (n − 1)-admissible and so by assumptionthere is an (M,n − 2, n − 1)-decomposition D of KU .

Let H be an (n − 1)-cycle in D, let C be an (n − 2)-cycle in D, let {I, I1} be a decompositionof C into two matchings, let [w, x, y, z] be a path in I1 ∪ H such that wx, yz ∈ E(I1) andxy ∈ E(H), and let v be the vertex in U \ V (C). Then

(D \ {C,H}) ∪ {I + v∞,H ′, (∞, w, x, y, z)} ∪ D1


(i) H ′ = (H − [x, y]) ∪ [x,∞, y]; and(ii) D1 is a (3(n−6)/2)-decomposition of K{∞},U\{v,w,x,y,z} ∪ (I − {wx, yz}).

5.3. Many 3-cycles, one Hamilton cycle and no 4- or 5-cycles

We show that the required decompositions exist in Lemma 36. We first require three preliminarylemmas. These results are proved using the edge swapping techniques mentioned previously.

Lemma 33. Let n and k be positive integers, and let M be a list of integers. If there existsan (M)-packing of Kn whose leave has a decomposition into two 3-cycles, T1 and T2, anda k-cycle C such that |V (T1) ∩ V (T2)| = 1, |V (T2) ∩ V (C)| = 1 and V (T1) ∩ V (C) = ∅, thenthere exists an (M, 3, k + 3)-decomposition of Kn.

Proof. Let P be an (M)-packing of Kn which satisfies the conditions of the lemma and let Lbe its leave. Let [w, x, y, z] be a path in L such that w ∈ V (T1) \ V (T2), V (T1) ∩ V (T2) = {x},V (T2) ∩ V (C) = {y}, and z ∈ V (C) \ V (T2). Let P ′ be the (M)-packing of Kn obtained fromP by performing the (w, z)-switch S with origin x. If the terminus of S is y, then the leave ofP ′ has a decomposition into the 3-cycle T2 and a (k + 3)-cycle. Otherwise the terminus of S is


not y and the leave of P ′ has a decomposition into the 3-cycle (x, y, z) and a (k + 3)-cycle. Ineither case we complete the proof by adding these cycles to P ′.

Lemma 34. Let n and k be positive integers such that k � n − 4, and let M be a list ofintegers. Suppose that there exists an (M)-packing of Kn whose leave has a decomposition intotwo 3-cycles, T1 and T2, and a k-cycle C such that |V (T1) ∩ V (C)| � 1, |V (T2) ∩ V (C)| � 1,and |V (T1) ∩ V (T2)| = 1 if k = n − 4. Then there exists an (M, 3, k + 3)-decomposition of Kn.

Proof. Let P be an (M)-packing of Kn which satisfies the conditions of the lemma and letL be its leave.

Case 1. Suppose that Δ(L) = 2. Then T1, T2 and C are pairwise vertex disjoint. Let x ∈V (T1) and y ∈ V (C) and let z be a neighbour in T1 of x. Let P ′ be the (M)-packing of Kn

obtained from P by performing the (x, y)-switch S with origin z, and let L′ be the leave of P ′.Then either the non-trivial components of L′ are a 3-cycle and a (k + 3)-cycle or Δ(L′) = 4,exactly one vertex of L′ has degree 4, and L′ has exactly two non-trivial components. In theformer case, we can add these cycles to P ′ to complete the proof. In the latter case we canapply Lemma 26 to complete the proof.

Case 2. Suppose that Δ(L) = 4. If exactly one vertex of L has degree 4, then L has exactlytwo non-trivial components and we can complete the proof by applying Lemma 26. Thus,we can assume that L has at least two vertices of degree 4. We can further assume that Pdoes not satisfy the conditions of Lemma 33, for otherwise we can complete the proof byapplying Lemma 33. Noting that |V (T1) ∩ V (T2)| ∈ {0, 1}, it follows that |V (T1) ∩ V (C)| = 1and |V (T2) ∩ V (C)| = 1. Because k � n − 4 and |V (T1) ∩ V (T2)| = 1 if k = n − 4, there is anisolated vertex z in L. Let w be the vertex in V (T1) ∩ V (C), let x and y be the neighbours in T1

of w. Let P ′ be the (M)-packing of Kn obtained from P by performing the (w, z)-switch S withorigin x, and let L′ be the leave of P ′. If the terminus of S is not y, then L′ has a decompositioninto a (k + 3)-cycle and a 3-cycle, and we complete the proof by adding these cycles to P ′.Otherwise the terminus of S is y and either Δ(L′) = 4, exactly one vertex of L′ has degree4, and L′ has exactly two non-trivial components (this occurs when |V (T1) ∩ V (T2)| = 0) orP ′ satisfies the conditions of Lemma 33 (this occurs when |V (T1) ∩ V (T2)| = 1). Thus, we cancomplete the proof by applying Lemma 26 or 33.

Case 3. Suppose that Δ(L) � 6. In this case, exactly one vertex of L has degree 6 and everyother vertex of L has degree at most 2. Let w be the vertex of degree 6 in L, let x be a neighbourin T2 of w, and let y and z be the neighbours in T1 of w. Let P ′ be the (M)-packing of Kn

obtained from P by performing the (w, x)-switch S with origin y, and let L′ be the leave ofP ′. If the terminus of S is z, then P ′ satisfies the conditions of Lemma 33 and we completethe proof by applying Lemma 33. Otherwise the terminus of S is not z and L′ has a decom-position into the 3-cycle T2 and a (k + 3)-cycle, and we complete the proof by adding thesecycles to P ′.

Lemma 35. Let n, k and t be positive integers such that 3 � k � n − 4. If there exists a(3t, k, n)-decomposition of Kn, then there exists a (3t−1, k + 3, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let V be a vertex set with |V | = n, letD be a (3t, k, n)-decomposition of KV , and let C be a k-cycle in D. Let U = V \ V (C). Then-cycle in D contains at most |U | − 1 edges of KU (as the subgraph of the n-cycle induced byU is a forest). Also, if n is even, then the perfect matching in D contains at most � 1

2 |U |� edgesof KU . The proof now splits into two cases depending on whether k = n − 4.


Case 1. Suppose that k � n − 5. Then |U | � 5 and, by the comments in the precedingparagraph, the 3-cycles in D contain at least four edges of KU . Thus, there are distinct 3-cycles T1, T2 ∈ D such that each contains at least one edge of KU . We can remove C, T1 andT2 from D and apply Lemma 34 to the resulting packing to complete the proof.

Case 2. Suppose that k = n − 4. Then |U | = 4, the n-cycle in D contains at most three edgesof KU and the perfect matching in D contains at most two edges of KU . This leaves at leastone edge of KU which occurs in a 3-cycle T1 ∈ D. Let U = {u1, u2, u3, u4} and let H be then-cycle in D. This case now splits into two subcases depending on whether V (T1) ∩ V (C) = ∅.

Case 2a. Suppose that V (T1) ∩ V (C) = ∅. Then we can assume without loss of generalitythat T1 = (u1, u2, u3). If any of the three edges u1u4, u2u4, u3u4 is in a 3-cycle T2 ∈ D, then wecan remove C, T1 and T2 from D and apply Lemma 34 to the resulting packing to completethe proof. Thus, we assume there is no such 3-cycle in D. Without loss of generality, it followsthat n is even, that u1u4 is an edge of the perfect matching in D, and that u2u4, u3u4 ∈ E(H).Let z be a vertex in C which is not adjacent in H to a vertex in U (such a vertex exists asn � 15 implies |V (C)| � 11 and there are only at most four vertices of C which are adjacentin H to vertices in U).

Now let P ′ be the (3t−1, n)-packing of KV obtained from D \ {C, T1} by performing the(u1, z)-switch S with origin u2. If the terminus of S is not u3, then the only non-trivialcomponent in the leave of P ′ is a (k + 3)-cycle and we can complete the proof by addingthis cycle to P ′. Otherwise, the terminus of S is u3 and the only non-trivial component in theleave of P ′ is (u2, u3, z) ∪ C. Furthermore, since z is not adjacent in H to a vertex in U , thefinal dot point in Lemma 24 guarantees that neither u1u2 nor u1u3 is an edge of the n-cyclein P ′. Since u1u2 and u1u3 cannot both be edges of the perfect matching in P ′, this meansthat one of them must be in a 3-cycle T ′

2 ∈ P ′. Thus, we can remove T ′2 from P ′ and apply

Lemma 34 to the resulting packing to complete the proof.Case 2b. Suppose that |V (T1) ∩ V (C)| = 1. Let T1 = (x, y, z) with x ∈ V (C) and y, z ∈ U ,

and let w ∈ U \ {y, z}. Let P ′ be the (3t−1, n)-packing of KV obtained from D \ {C, T1} byperforming the (w, x)-switch S with origin y, and let L′ be the leave of P ′. If the terminus ofS is not z, then the only non-trivial component in the leave of P ′ is a (k + 3)-cycle and we cancomplete the proof by adding this cycle to P ′. Otherwise the terminus of S is z, and the onlynon-trivial components in the leave of P ′ are C and (w, y, z). By adding these cycles to P ′ weobtain a (3t, k, n)-decomposition of KV which contains a 3-cycle and an (n − 4)-cycle whichare vertex disjoint, and we can proceed as we did in Case 2a.

We are now ready to prove the main result of this subsection.

Lemma 36. If n, k and t are positive integers such that 3 � k � n − 1, Theorem 1holds for Kn−3, Kn−2 and Kn−1, and (3t, k, n) is an n-ancestor list, then there is a(3t, k, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let r ∈ {3, 4, 5} such that r ≡ k (mod 3).It suffices to find a (3t+(k−r)/3, r, n)-decomposition of Kn, since we can then obtain a (3t, k, n)-decomposition of Kn by repeatedly applying Lemma 35 ((k − r)/3 times). If r = 3, then theexistence of a (3t+(k−r)/3, r, n)-decomposition of Kn follows from the main result of [25], so wemay assume r ∈ {4, 5}. Thus, the existence of the required (3t+(k−r)/3, r, n)-decompositionof Kn is given by one of Lemmas 29–32, provided that t + (k − r)/3 � (3n − 14)/2 (thenumber of 3-cycles in the decompositions given by Lemma 31 is at least (3n − 14)/2 andthe number is smaller for the other three lemmas for n � 15). However, it follows from3t + k + n = n�(n − 1)/2�, k � n − 1 and n � 15 that t � (3n − 14)/2.


5.4. Many 4-cycles and one Hamilton cycle

In Lemma 37 we construct the required decompositions of complete graphs of odd order, and inLemma 38 we construct the required decompositions of complete graphs of even order. In eachcase we construct the required decomposition of Kn from a suitable decomposition of Kn−2.

Lemma 37. If n is odd, Theorem 1 holds for Kn−2, and (M, 4(n−3)/2, n) is an n-ancestorlist with νn(M) = 0, then there is an (M, 4(n−3)/2, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 15. Let U be a vertex set with |U | = n − 2, let∞1 and ∞2 be distinct vertices not in U , and let V = U ∪ {∞1,∞2}. Since (M, 4(n−3)/2, n) isan n-ancestor list with νn(M) = 0, it follows from (5) in the definition of ancestor lists that anycycle length in M is at most n − 3. Thus, it is easily seen that (M,n − 3) is (n − 2)-admissibleand so by assumption there is an (M,n − 3)-decomposition D of KU .

Let H be an (n − 3)-cycle in D, let z be the vertex in U \ V (H), and let x and y be adjacentvertices in H. Then

(D \ {H}) ∪ {H ′, (∞1, y, x,∞2)} ∪ D1

is an (M, 4(n−3)/2, n)-decomposition of KV , where

(i) H ′ = (H − [x, y]) ∪ [x,∞1, z,∞2, y]; and(ii) D1 is a (4(n−5)/2)-decomposition of K{∞1,∞2},U\{x,y,z}.

Lemma 38. If n is even, Theorem 1 holds for Kn−2, and (M, 4(n−2)/2, n) is an n-ancestorlist with νn(M) = 0, then there is an (M, 4(n−2)/2, n)-decomposition of Kn.

Proof. By Lemma 2, we can assume that n � 16. Let U be a vertex set with |U | = n − 2, let∞1 and ∞2 be distinct vertices not in U , and let V = U ∪ {∞1,∞2}. Since (M, 4(n−2)/2, n) isan n-ancestor list with νn(M) = 0, it follows from (5) in the definition of ancestor lists that anycycle length in M is at most n − 3. Thus, it is easily seen that (M, 3, n − 3) is (n − 2)-admissibleand so by assumption there is an (M, 3, n − 3)-decomposition D of KU .

Let H be an (n − 3)-cycle in D, let C be a 3-cycle in D, and let I be the perfect matching inD. Let z be the vertex in U \ V (H), let w and x be distinct vertices in V (C) ∩ V (H), let u bethe vertex in V (C) \ {w, x} (possibly u = z), and let y be a vertex adjacent to x in H. Then

(D \ {I, C,H}) ∪ {I + ∞1∞2,H′, (∞1, y, x, w), (∞2, x, u, w)} ∪ D1

is an (M, 4(n−2)/2, n)-decomposition of KV , where

• H ′ = (H − [x, y]) ∪ [x,∞1, z,∞2, y]; and• D1 is a (4(n−6)/2)-decomposition of K{∞1,∞2},U\{w,x,y,z}.

5.5. Many 5-cycles and one Hamilton cycle

In Lemma 43 we construct the required decompositions of complete graphs of odd order, andin Lemma 44 we construct the required decompositions of complete graphs of even order. Wefirst require four preliminary lemmas.

Lemma 39. Every even graph has a decomposition into cycles such that any two cycles inthe decomposition share at most two vertices.


Proof. It is well known that every even graph has a decomposition into cycles. Let G be aneven graph. Amongst all decompositions of G into cycles, let D be one with a maximum numberof cycles. We claim that any pair of cycles in D shares at most two vertices. Suppose otherwise.That is, there are distinct cycles A and B in D and distinct vertices x, y and z of G such that{x, y, z} ⊆ V (A) ∩ V (B). Let A = A1 ∪ A2 and B = B1 ∪ B2, where A1, A2, B1 and B2 arepaths from x to y such that z ∈ V (A2) and z ∈ V (B2). Then it is easy to see that A1 ∪ B1 andA2 ∪ B2 are both non-empty even graphs. For i = 1, 2, let Di be a decomposition of Ai ∪ Bi

into cycles, and note that |D2| � 2 because degA2∪B2(z) = 4. Then (D \ {A,B}) ∪ D1 ∪ D2

is a decomposition of G into cycles which contains more cycles than D, contradicting ourdefinition of D.

Lemma 40. Let V be a vertex set and let U be a subset of V such that |U | � 10. Let Mbe a list of integers, let m ∈ {3, 4, 5}, and let P be an (M)-packing of KV with a leave L suchthat degL(x) = 0 for all x ∈ V \ U . If there exists a spanning even subgraph G of L such thatat least one of the following holds:

(i) Δ(G) = 4, exactly one vertex of G has degree 4, G has at most two non-trivialcomponents, |E(G)| � m + 3, and G does not have a decomposition into two odd cyclesif m = 4;

(ii) Δ(G) = 4, exactly two vertices of G have degree 4, G has exactly one non-trivialcomponent, |E(G)| � m + 3, and degG(x) = 0 for some x ∈ U if m = 3;

(iii) m = 4, G has exactly one non-trivial component, G has a decomposition into threecycles each pair of which intersect in exactly one vertex, and degG(x) = 0 for somex ∈ U ; or

(iv) m = 5, Δ(G) � 4, and G has a decomposition into three cycles such that any twointersect in at most two vertices, and such that any two which intersect have lengthsadding to 6 or 7;

then there exists (M,m)-packing P ′ of KV which is equivalent to P on V \ U .

Proof. Suppose that there is a spanning even subgraph G of L which satisfies one of (i),(ii), (iii) or (iv). Because L and G are both even graphs, it follows that L − G is an even graphand hence has a decomposition A = {A1, . . . , At} into cycles. Let ai = |V (Ai)| for i = 1, . . . , tand let M† = (a1, . . . , at). So P ∪ A is an (M,M†)-packing of KV which is equivalent to P onV \ U . The leave of P ∪ A is G. Let e = |E(G)|.

If we can produce an (M,M†,m, e − m)-decomposition D of KV which is equivalent to Pon V \ U , then there will be cycles in D with lengths a1, . . . , at, e − m whose vertex sets aresubsets of U , and we can complete the proof by removing these cycles from D. So, it sufficesto find such a decomposition. The proof now splits into cases.

Case 1. Suppose that G satisfies (i). Then we can apply Lemma 26 to obtain the requireddecomposition.

Case 2. Suppose that G satisfies (ii). Then we can apply Lemma 27 to obtain the requireddecomposition. The only non-trivial thing to check is that there is an x ∈ U with degG(x) = 0when m ∈ {4, 5} and e − m = 3. In this case we have e ∈ {7, 8} and, because G is even and|U | � 10, there is indeed an x ∈ U with degG(x) = 0.

Case 3. Suppose that G satisfies (iii). Either exactly one vertex of G has degree 6 and everyother vertex of G has degree at most 2, or exactly three vertices of G have degree 4 and everyother vertex of G has degree at most 2. In the former case, we apply Lemma 25, choosing bto be the vertex of degree 6 in G and a to be a neighbour in G of b. In the latter case weapply Lemma 25, choosing b to be a vertex of degree 4 in G and a to be a vertex in U whichhas degree 0 in G. In either case we obtain an (M,M†)-packing P ′ of KV , which is equivalent


to P on V \ U , with a leave G′ such that Δ(G′) = 4, exactly two vertices of G′ have degree 4,G′ has exactly one non-trivial component, and |E(G′)| � 9. Thus, we can apply Lemma 27 toobtain the required decomposition.

Case 4. Suppose that G satisfies (iv). Since Δ(G) � 4 there is at least one pair of intersectingcycles in any cycle decomposition of G. Thus, there exists a decomposition {B1, B2, B3} of Ginto three cycles such that |V (B1) ∩ V (B2)| ∈ {1, 2} and |E(B1)| + |E(B2)| ∈ {6, 7}.

Case 4a. Suppose that B3 is a component of G. If |V (B1) ∩ V (B2)| = 1, then we can applyLemma 26 to obtain the required decomposition, so we may assume that |V (B1) ∩ V (B2)| = 2.Let x ∈ V (B1) ∩ V (B2), let y ∈ V (B3), and let z be a neighbour in G of x. Let P ′ be the(M,M†)-packing of Kn obtained from P by performing the (x, y)-switch S with origin z and letG′ be its leave. Then P ′ is equivalent to P on V \ U , G′ has exactly one non-trivial component,Δ(G′) = 4, exactly two vertices of G′ have degree 4, and |E(G′)| � 9. Thus, we can applyLemma 27 to obtain the required decomposition.

Case 4b. Suppose that B3 is not a component of G. Then B3 intersects with B1 or B2

and so |E(B3)| ∈ {3, 4}. Thus, e ∈ {9, 10, 11} as we have |E(B1)| + |E(B2)| ∈ {6, 7}. Let P ′

be the (M,M†)-packing of Kn obtained from P by repeatedly applying Lemma 25, each timechoosing b to be a vertex maximum degree in the leave and a to be a vertex in U of degree0 in the leave, until the leave has maximum degree 4 and has exactly one vertex of degree 4(a suitable choice for a will exist each time since e � 11 and |U | � 10). Let G′ be the leave ofP ′. Then P ′ is equivalent to P on V \ U , Δ(G′) = 4, exactly one vertex of G′ has degree 4,|E(G′)| ∈ {9, 10, 11}, and G′ has at most two components (because |E(G′)| � 11). Thus, wecan apply Lemma 26 to obtain the required decomposition.

Lemma 41. Let V be a vertex set and let U be a subset of V such that |U | � 10. Letm ∈ {3, 4, 5}, let M be a list of integers, and let P be an (M)-packing of KV with a leave L suchthat |E(L)| � |U | + m and degL(x) = 0 for all x ∈ V \ U . Then there exists an (M,m)-packingof KV which is equivalent to P on V \ U .

Proof. Since L is an even graph, Lemma 39 guarantees that there is a decomposition Dof L such that any pair of cycles in D intersect in at most two vertices. Let e = |E(L)|. Sincee � |U | + m � 13 it follows that D contains at least three cycles. Also, since e > |U |, there is atleast one pair of intersecting cycles in D. We now consider separately the cases m = 3, m = 4and m = 5.

Case 1. Suppose that m = 3. We can assume that there are no 3-cycles in D (otherwise wecan simply add one to P to complete the proof). Let C1, C2 and C3 be distinct cycles in Dsuch that C1 and C2 intersect. If |V (C1) ∩ V (C2)| = 1, then we can apply Lemma 40(i) (withE(G) = E(C1 ∪ C2)) to complete the proof, so we may assume that |V (C1) ∩ V (C2)| = 2. If|E(C1)| + |E(C2)| � |U | + 1, then there is at least one vertex of U that is not in V (C1) ∪ V (C2)and we can apply Lemma 40(ii) (with E(G) = E(C1 ∪ C2)) to complete the proof. Thus, we mayassume |E(C1)| + |E(C2)| = |U | + 2, and it follows from this that V (C1) ∪ V (C2) = U . Thismeans that V (C3) ⊆ V (C1) ∪ V (C2). Thus, since we have V (C3) � 4, V (C1) ∩ V (C3) � 2 andV (C2) ∩ V (C3) � 2, it follows that V (C3) = 4, V (C1) ∩ V (C3) = 2 and V (C2) ∩ V (C3) = 2.We can assume, without loss of generality, that |E(C1)| � |E(C2)| and hence that |E(C2)| � 5(since |U | � 10). This means that there is at least one vertex of U that is in neither C1 norC3, and so we can apply Lemma 40 (ii) (with E(G) = E(C1 ∪ C3)) to complete the proof.

Case 2. Suppose that m = 4. If two cycles in D intersect in exactly two vertices, then wecan apply Lemma 40(ii) (with the edges of G being the edges of two such cycles) to completethe proof. So, we may assume that any two cycles in D intersect in at most one vertex. Let{C1, C2} be a pair of intersecting cycles in D such that |E(C1 ∪ C2)| � |E(Ci ∪ Cj)| for anypair {Ci, Cj} of intersecting cycles in D. If there is a cycle in D which is vertex disjoint from


C1 ∪ C2, then we can apply Lemma 40(i) (with the edges of G being the edges of C1, C2 andthis cycle) to complete the proof. If there is a cycle in D which intersects with exactly oneof C1 and C2, then we can apply Lemma 40(ii) (with the edges of G being the edges of C1,C2 and this cycle) to complete the proof. So, we may assume that every cycle in D \ {C1, C2}intersects (in exactly one vertex) with C1 and with C2. Let C3 be a shortest cycle in D \{C1, C2} and note that |V (Ci)| � |V (C3)| for i = 1, 2 by our definition of C1 and C2. If V (C1 ∪C2 ∪ C3) = U , then we can apply Lemma 40(iii) (with E(G) = E(C1 ∪ C2 ∪ C3)) to completethe proof. Otherwise V (C1 ∪ C2 ∪ C3) = U which means that |V (C1)| + |V (C2)| + |V (C3)| ∈{|U | + 2, |U | + 3}. However, we have e � |U | + 4 and so there is a cycle C4 ∈ D \ {C1, C2, C3}.Thus, C4 is a 3-cycle (as V (C1 ∪ C2 ∪ C3) = U and C4 intersects each of C1, C2 and C3 inexactly one vertex). It then follows from the minimality of C3 and from |V (Ci)| � |V (C3)| fori = 1, 2 that C1, C2 and C3 are also 3-cycles. Since |U | � 10, this is a contradiction and theresult is proved.

Case 3. Suppose that m = 5. Let C1, C2 and C3 be three cycles in D such that C1 and C2

intersect. If there are a pair of cycles in {C1, C2, C3} which intersect and whose lengths addto at least 8, then the union of this pair of cycles has at least m + 3 edges and we can applyLemma 40(i) or Lemma 40(ii) (with the edges of G being the edges of this pair of cycles) tocomplete the proof. Otherwise we can apply Lemma 40(iv) to complete the proof.

Lemma 42. Let u and k be integers such that u is even, u � 16 and 6 � k � u − 1, let Ube a vertex set such that |U | = u, and let x and y be distinct vertices in U . Then there exists apacking of KU with a perfect matching, a u-cycle, a (u − 1)-path from x to y, a k-cycle, three(u − 2)-cycles each having vertex set U \ {x, y}, and a 2-path from x to y.

Proof. Let U = Zu−3 ∪ {∞, x, y}. For i = 0, . . . , 5, let

Hi =(∞, i, i + 1, i + (u − 4), i + 2, i + (u − 5), . . . , i +

u − 62

, i +u

2, i +

u − 42

, i +u − 2

2

),

and let

I =[u − 6

2,u − 2

2

]∪

[u − 8

2,u

2

]∪ · · · ∪ [0, u − 4] ∪

[∞,

u − 42

],

so that {I,H0, . . . , H5} is a packing of KZu−3∪{∞} with one perfect matching and six (u −2)-cycles (recall that u � 16). Then

{I + xy, (H0 − [∞, 0, 1]) ∪ [∞, x, 0, y, 1], (H1 − [1, 2]) ∪ [1, x] ∪ [2, y],P ∪ [a, x, b],H3,H4,H5, [x, c, y]}

is the required packing, where P is a (k − 2)-path in H2 with endpoints a and b such thata, b ∈ Zu−3 \ {0, 1} (P exists as there are u − 2 distinct paths of length k − 2 in H2, and atmost six having ∞, 0 or 1 as an endpoint), and c is any vertex in Zu−3 \ {0, 1, 2, a, b}.

Lemma 43. If n is odd and (M, 5(3n−11)/2, n) is an n-ancestor list with νn(M) = 0, thenthere is an (M, 5(3n−11)/2, n)-decomposition of Kn.

Proof. By Lemma 2 (for n � 13) and Lemma 28 (for n ∈ {15, 17}), we may assume thatn � 19 (Lemma 28 can indeed be applied as ν5(M, 5(3n−11)/2, n) � 3 when n ∈ {15, 17}). LetU be a vertex set with |U | = n − 3, let x and y be distinct vertices in U , let ∞†, ∞1 and ∞2

be distinct vertices not in U , and let V = U ∪ {∞†,∞1,∞2}.Since (M, 5(3n−11)/2, n) is an n-ancestor list with νn(M) = 0, it follows from (6) in the

definition of ancestor lists that any cycle length in M is at most n − 5. If there is a cycle


length in M which is at least 6, then let k be this cycle length. Otherwise let k = 0 (so k ∈{0} ∪ {6, . . . , n − 5}). By Lemma 42, there exists a packing P of KU with a perfect matchingI, an (n − 4)-path P1 from x to y, a k-cycle (if k = 0), three (n − 5)-cycles C1, C2 and C3 eachhaving vertex set U \ {x, y}, and a 2-path P2 from x to y. Let {I1, I2} be a decomposition ofC3 into two matchings.

Let

P ′ = (P \ {I, P1, P2, C1, C2, C3}) ∪ {P1 ∪ [x,∞1,∞†,∞2, y],P2 ∪ [x,∞2,∞1, y]} ∪ D ∪ D1 ∪ D2,

where

(i) D is a (3(n−3)/2)-decomposition of K{∞†} ∨ I;(ii) for i = 1, 2, Di is a (5(n−5)/2)-decomposition of K{∞i} ∨ (Ci ∪ Ii) (this exists by

Lemma 20).

Then P ′ is a (3(n−3)/2, 5n−4, k, n)-packing of KV (a (3(n−3)/2, 5n−4, n)-packing of KV if k = 0)such that

(i) (n − 3)/2 3-cycles in P ′ contain the vertex ∞†;(ii) ∞†∞1 and ∞†∞2 are edges of the n-cycle in P ′; and(iii) ∞†, ∞1 and ∞2 all have degree 0 in the leave of P ′.

Since (M, 5(3n−11)/2, n) is an n-ancestor list with νn(M) = 0, it can be seen that by beginningwith P ′ and repeatedly applying Lemma 41 we can obtain an (M, 3(n−3)/2, 5n−4, n)-packingof KV which is equivalent to P ′ on V \ U . Note that the leave of this packing has n − 3edges. Thus, by then repeatedly applying Lemma 25 we can obtain an (M, 3(n−3)/2, 5n−4, n)-packing P ′′ of KV which is equivalent to P ′ on V \ U and whose leave L′′ has the propertythat degL′′(x) = 0 for each x ∈ {∞†,∞1,∞2} and degL′′(x) = 2 for each x ∈ U . Because P ′′

is equivalent to P ′ on V \ U , it follows from (i) and (ii) that there is a set T of n−32 3-cycles

in P ′′ each of which contains the vertex ∞† and two vertices in U . Let T be the union of the3-cycles in T . Then

(P ′′ \ T ) ∪ D′′

is an (M, 5(3n−11)/2, n)-decomposition of KV where D′′ is a (5(n−3)/2)-decomposition of T ∪ L′′

(this exists by Lemma 20, noting that E(T ∪ L′′) = E(K{∞†} ∨ G) for some 3-regular graphG on vertex set U which contains a perfect matching).

Lemma 44. If n is even and (M, 52n−9, n) is an n-ancestor list with νn(M) = 0, then thereis an (M, 52n−9, n)-decomposition of Kn.

Proof. By Lemma 2 (for n � 14) and Lemma 28 (for n ∈ {16, 18}), we may assume thatn � 20 (Lemma 28 can indeed be applied as ν5(M, 52n−9, n) � 3 when n ∈ {16, 18}). Let U bea vertex set with |U | = n − 4, let x and y be distinct vertices in U , let ∞†

1, ∞†2, ∞1 and ∞2

be distinct vertices not in U , and let V = U ∪ {∞†1,∞

†2,∞1,∞2}.

Since (M, 52n−9, n) is an n-ancestor list with νn(M) = 0, it follows from (6) in the definitionof ancestor lists that any cycle length in M is at most n − 5. If there is a cycle length in M whichis at least 6 then let k be this cycle length. Otherwise let k = 0 (so k ∈ {0} ∪ {6, . . . , n − 5}).By Lemma 42, there exists a packing P of KU with a perfect matching I, an (n − 4)-cycle B,an (n − 5)-path P1 from x to y, a k-cycle (if k = 0), and three (n − 6)-cycles C1, C2 and C3

each having vertex set U \ {x, y}, and a 2-path P2 from x to y. Let {I†1 , I†2} be a decompositionof B into two matchings and {I1, I2} be a decomposition of C3 into two matchings.


Let

P ′ = (P \ {B,P1, P2, C1, C2, C3}) ∪ {I + {∞1∞†2,∞2∞†

1},P1 ∪ [x,∞1,∞†

1,∞†2,∞2, y],

P2 ∪ [x,∞2,∞1, y]} ∪ D†1 ∪ D†

2 ∪ D1 ∪ D2,

where

(i) for i = 1, 2, D†i is a (3(n−4)/2)-decomposition of K{∞†

i} ∨ I†i ;(ii) for i = 1, 2, Di is a (5(n−6)/2)-decomposition of K{∞i} ∨ (Ci ∪ Ii) (this exists by

Lemma 20).

Then P ′ is a (3n−4, 5n−5, k, n)-packing of KV (a (3n−4, 5n−5, n)-packing of KV if k = 0) suchthat

(i) for i = 1, 2, (n − 4)/2 3-cycles in P ′ contain the vertex ∞†i ;

(ii) ∞†1∞1, ∞†

1∞†2 and ∞†

2∞2 are edges of the n-cycle in P ′;

(iii) ∞†1∞2 and ∞†

2∞1 are edges of the perfect matching in P ′; and

(iv) ∞†1, ∞

†2, ∞1 and ∞2 all have degree 0 in the leave of P ′.

Since (M, 52n−9, n) is an n-ancestor list with νn(M) = 0, by beginning with P ′ andrepeatedly applying Lemma 41 we can obtain an (M, 3n−4, 5n−5, n)-packing of KV , whichis equivalent to P ′ on V \ U . Note that the leave of this packing has 2n − 8 edges. Thus, bythen repeatedly applying Lemma 25 we can obtain an (M, 3n−4, 5n−5, n)-packing P ′′ of KV

which is equivalent to P ′ on V \ U and whose leave L′′ has the property that degL′′(x) = 0 forx ∈ {∞†

1,∞†2,∞1,∞2} and degL′′(x) = 4 for all x ∈ U . By Petersen’s Theorem [41], L′′ has

a decomposition {H1,H2} into two 2-regular graphs, each with vertex set U . Because P ′′ isequivalent to P ′ on V \ U , it follows from (i), (ii) and (iii) that, for i = 1, 2 there is a set Ti

of (n − 4)/2 3-cycles in P ′′ each of which contains the vertex ∞†i and two vertices in U . For

i = 1, 2, let Ti be the union of the 3-cycles in Ti. Then

(P ′′ \ (T1 ∪ T2)) ∪ D′′1 ∪ D′′

2

is an (M, 52n−9, n)-decomposition of KV where, for i = 1, 2, D′′i is a (5(n−4)/2)-decomposition

of Ti ∪ Hi (these decompositions exist by Lemma 20, noting that for i = 1, 2, E(Ti ∪ Hi) =E(K{∞†

i} ∨ G) for some 3-regular graph G with vertex set U that contains a perfectmatching).

5.6. Proof of Lemma 5 in the case of one Hamilton cycle

Lemma 45. If Theorem 1 holds for Kn−1, Kn−2 and Kn−3, then there is an (M)-decomposition of Kn for each n-ancestor list M satisfying νn(M) = 1.

Proof. By Lemma 2, we can assume that n � 15. If there is a cycle length in M which isat least 6 and at most n − 1, then let k be this cycle length. Otherwise let k = 0. We dealseparately with the case n is odd and the case n is even.

Case 1. Suppose that n is odd. Since n � 15 and 3ν3(M) + 4ν4(M) + 5ν5(M) + k + n =n(n−1)

2 , it can be seen that either

(i) n ∈ {15, 17, 19} and ν5(M) � 3;(ii) ν3(M) � (n − 5)/2;(iii) ν4(M) � (n − 3)/2; or(iv) ν5(M) � (3n − 11)/2.


(To see this consider the cases ν5(M) � 3 and ν5(M) � 2 separately and use the definition ofn-ancestor list.) If (i) holds, then the result follows by Lemma 28. If (ii) holds, then the resultfollows by one of Lemma 29, 30 or 36. If (iii) holds, then the result follows by Lemma 37. If(iv) holds, then the result follows by Lemma 43.

Case 2. Suppose that n is even. Since n � 16 and 3ν3(M) + 4ν4(M) + 5ν5(M) + k + n =n(n − 2)/2, it can be seen that either

(i) n ∈ {16, 18, 20, 22, 24, 26}, ν5(M) � 3, and ν4(M) � 2 if n = 24;(ii) ν3(M) � (3n − 14)/2;(iii) ν4(M) � (n − 2)/2; or(iv) ν5(M) � 2n − 9.

(To see this consider the cases ν5(M) � 3 and ν5(M) � 2 separately and use the definitionof n-ancestor list.) If (i) holds, then the result follows by Lemma 28. If (ii) holds, then theresult follows by one of Lemma 31, 32 or 36 (note that (3n − 14)/2 � (n − 6)/2 for n � 16).If (iii) holds, then the result follows by Lemma 38. If (iv) holds, then the result follows byLemma 44.

6. Decompositions of 〈S〉nOur general approach to constructing decompositions of 〈S〉n follows the approach usedin [27, 29]. For each connection set S in which we are interested, we define a graph Jn for eachpositive integer n such that there is a natural bijection between E(Jn) and E(〈S〉n), and suchthat 〈S〉n can be obtained from Jn by identifying a small number (approximately |S|) of pairsof vertices. Thus, decompositions of Jn yield decompositions of 〈S〉n.

The key property of the graph Jn is that it can be decomposed into a copy of Jn−y and acopy of Jy for any positive integer y such that 1 � y < n, and this facilitates the constructionof desired decompositions of Jn for arbitrarily large n from decompositions of Ji for varioussmall values of i. For example, in the case S = {1, 2, 3} we define Jn by V (Jn) = {0, . . . , n +2} and E(Jn) = {{i, i + 1}, {i + 1, i + 3}, {i, i + 3}} : i = 0, . . . n − 1}. It is straightforward toconstruct a (3)-decomposition of J1, a (4, 5)-decomposition of J3, a (43)-decomposition of J4

and a (53)-decomposition of J5. Moreover, since Jn decomposes into Jn−y and Jy, it is easyto see that these decompositions can be combined to produce an (M)-decomposition of Jn

for any list M = (m1, . . . ,mt) satisfying∑

M = 3n and mi ∈ {3, 4, 5} for i = 1, . . . , t. For alln � 7, an (M)-decomposition of 〈{1, 2, 3}〉n can be obtained from an (M)-decomposition of Jn

by identifying vertex i with vertex i + n for i = 0, 1, 2.In what follows, this general approach is modified to allow for the construction of decompo-

sitions which, in addition to cycles of lengths 3, 4 and 5, contain one arbitrarily long cycle or,in the case S = {1, 2, 3}, one arbitrarily long cycle and one Hamilton cycle. The constructionsused to prove Lemma 7 proceed in a similar fashion for each connection set S. We include theproof only for the most interesting and difficult case, namely the case S = {1, 2, 3, 4, 6}, wherethere are some exceptional decompositions which appear not to exist. Many decompositions ofvarious Jn with n small are required as ingredients in our constructions. All of these ingredientshave been constructed, in most cases with the aid of a computer, but due to the large numberof them, they are not presented here.

6.1. Proof of Lemma 7

In this section we prove Lemma 7, which we restate here for convenience.


Lemma (Lemma 7). If

S ∈ {{1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7},{1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7, 8}},

n � 2max(S) + 1, and M = (m1, . . . ,mt, k) is any list satisfying mi ∈ {3, 4, 5} for i = 1, . . . , t,3 � k � n, and

∑M = |S|n, then there is an (M)-decomposition of 〈S〉n, except possibly

when

(i) S = {1, 2, 3, 4, 6}, n ≡ 3 (mod 6) and M = (35n/3); or(ii) S = {1, 2, 3, 4, 6}, n ≡ 4 (mod 6) and M = (3(5n−5)/3, 5).

Let

S = {{1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 5, 7},{1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7, 8}}.

For n � 1 and for each S ∈ S, we define the graph JSn according to the following table. We

shall give the proof of Lemma 7 for S = {1, 2, 3, 4, 6} only. Thus, we shall be working with thegraph J

{1,2,3,4,6}n , which in this subsection we denote by just Jn. For each other S ∈ S, we use

the graph JSn given in the table, and then the proof proceeds along similar lines.

S V (JSn ) E(JS

n ){1, 2, 3} {0, . . . , n + 2} {{i, i + 1}, {i + 1, i + 3},

{i, i + 3} : i = 0, . . . , n − 1}{1, 2, 3, 4} {0, . . . , n + 3} {{i + 2, i + 3}, {i + 2, i + 4}, {i, i + 3},

{i, i + 4} : i = 0, . . . , n − 1}{1, 2, 3, 4, 6} {0, . . . , n + 5} {{i + 2, i + 3}, {i + 2, i + 4}, {i + 3, i + 6}, {i, i + 4},

{i, i + 6} : i = 0, . . . , n − 1}{1, 2, 3, 4, 5, 7} {0, . . . , n + 6} {{i + 6, i + 7}, {i + 5, i + 7}, {i + 3, i + 6}, {i + 3, i + 7},

{i, i + 5}, {i, i + 7} : i = 0, . . . , n − 1}{1, 2, 3, 4, 5, 6, 7} {0, . . . , n + 6} {{i + 3, i + 4}, {i + 5, i + 7}, {i + 3, i + 6}, {i, i + 4},

{i, i + 5}, {i, i + 6}, {i, i + 7} : i = 0, . . . , n − 1}{1, 2, 3, 4, 5, 6, 7, 8} {0, . . . , n + 8} {{i + 7, i + 8}, {i + 5, i + 7}, {i + 5, i + 8}, {i + 5, i + 9},

{i, i + 5}, {i + 1, i + 7}, {i, i + 7},{i + 1, i + 9} : i = 0, . . . , n − 1}

For a list of integers M , an (M)-decomposition of Jn will be denoted by Jn → M . We notethe following basic properties of Jn.

(i) For n � 13, if for each i ∈ {0, 1, 2, 3, 4, 5} we identify vertex i of Jn with vertex i + n ofJn then the resulting graph is 〈{1, 2, 3, 4, 6}〉n. This means that for n � 13, we can obtain an(M)-decomposition of 〈{1, 2, 3, 4, 6}〉n from a decomposition Jn → M , provided that for eachi ∈ {0, 1, 2, 3, 4, 5}, no cycle in the decomposition of Jn contains both vertex i and vertex i + n.

(ii) For any integers y and n such that 1 � y < n, the graph Jn is the union of Jn−y andthe graph obtained from Jy by applying the permutation x �→ x + (n − y). Thus, if there is adecomposition Jn−y → M and a decomposition Jy → M ′, then there is a decomposition Jn →M,M ′. We will call this construction, and the similar constructions that follow, concatenations.

We prove Lemma 7 by constructing various decompositions of graphs Jn and related graphs,then concatenating these decompositions to produce a much wider range of decompositions of


graphs Jn, and finally identifying pairs of vertices to produce the decompositions of 〈S〉n thatwe require.

Lemma 46. If n is a positive integer and M = (m1, . . . ,mt) is a list such that∑

M = 5n,mi ∈ {3, 4, 5} for i = 1, . . . , t, and M /∈ E where

E = {(32, 4)} ∪ {(35i) : i � 1 is odd} ∪ {(35i, 5) : i � 1 is odd},

then there is a decomposition Jn → M .

Proof. We have verified by computer search and concatenation that the result holds forn � 10. So, assume n � 11 and suppose by induction that the result holds for each integer n′

in the range 1 � n′ < n. It is routine to check that if M satisfies the hypotheses of the lemma,then M can be written as M = (X,Y ) where Jy → Y is one of the six decompositions below(which exist because y � 10).

J1 −→ 5, J4 −→ 45, J6 −→ 310, J5 −→ 37, 4, J4 −→ 34, 42, J3 −→ 3, 43.

If X /∈ E , then we can obtain a decomposition Jn → M by concatenating a decompositionJn−y → X (which exists by our inductive hypothesis) with a decomposition Jy → Y . Thus,we can assume X ∈ E . But since n � 11 and

∑Y � 30, we have

∑X � 25 which implies

X ∈ {(35i) : i � 3 is odd} ∪ {(35i, 5) : i � 3 is odd}. It follows that M = (310,X ′) for somenon-empty list X ′ /∈ E (because M /∈ E) and we can obtain a decomposition Jn → M byconcatenating a decomposition Jn−6 → X ′ (which exists by our inductive hypothesis) witha decomposition J6 → 310.

Let M = (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t. A decomposition{G1, . . . , Gt, C} of Jn such that

(i) Gi is an mi-cycle for i = 1, . . . , t; and(ii) C is a k-cycle such that V (C) = {n − k + 6, . . . , n + 5} and {{n, n + 3}, {n + 1, n +

4}, {n + 2, n + 5}} ⊆ E(C);

will be denoted Jn → M,k∗.In Lemma 47, we will form new decompositions of graphs Jn by concatenating decomposi-

tions of Jn−y with decompositions of graphs J+y which we will now define. For y ∈ {4, . . . , 11},

the graph obtained from Jy by adding the three edges {0, 3}, {1, 4} and {2, 5} will be denotedJ+

y . Let M = (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t. A decomposition{G1, . . . , Gt, A1, A2, A3} of J+

y such that

(i) Gi is an mi-cycle for i = 1, . . . , t;(ii) A1, A2 and A3 are vertex-disjoint paths, one from 0 to 3, one from 1 to 4, and one from

2 to 5; and(iii) |E(A1)| + |E(A2)| + |E(A3)| = l + 3

will be denoted J+y → M, l+. Moreover, if l = y and {{n, n + 3}, {n + 1, n + 4}, {n + 2, n +

5}} ⊆ E(A1) ∪ E(A2) ∪ E(A3), then the decomposition will be denoted J+y → M,y+∗.

For y ∈ {4, . . . , 11} and n > y, the graph Jn is the union of the graph obtained from Jn−y

by deleting the edges in {{n − y, n − y + 3}, {n − y + 1, n − y + 4}, {n − y + 2, n − y + 5}}and the graph obtained from J+

y applying the permutation x �→ x + (n − y). It followsthat if there is a decomposition Jn−y → M,k∗ and a decomposition J+

y → M ′, l+, thenthere is a decomposition Jn → M,M ′, k + l. The three edges {n, n + 3}, {n + 1, n + 4} and{n + 2, n + 5} of the k-cycle in the decomposition of Jn−y are replaced by the three paths


in the decomposition of J+y to form the (k + l)-cycle in the new decomposition. Similarly, if

there is a decomposition Jn−y → M,k∗ and a decomposition J+y → M ′, y+∗, then there is a

decomposition Jn → M,M ′, (k + y)∗.

Lemma 47. If n � 11 is an integer and M = (m1, . . . ,mt) is a list such that∑

M = 4nand mi ∈ {3, 4, 5} for i = 1, . . . , t, then there is a decomposition Jn → M,n∗.

Proof. We have verified using computer search and concatenation that the result holds for11 � n � 16. So, let n � 17 and suppose by induction that the result holds for each integer n′

in the range 11 � n′ < n. We have verified that the following decompositions exist:

J+4 −→ 44, 4+∗, J+

5 −→ 54, 5+∗, J+6 −→ 38, 6+∗.

It is routine to check, using∑

M = 4n � 68, that M can be written as M = (X,Y ) whereJ+

y → Y, y+∗ is one of the decompositions above and X is some non-empty list. We canobtain a decomposition Jn → M,n∗ by concatenating a decomposition Jn−y → X, (n − y)∗

(which exists by our inductive hypothesis, since n − y � n − 6 � 11) with a decompositionJ+

y → Y, y+∗.

Lemma 48. If n and k are integers such that 6 � k � n and M = (m1, . . . ,mt) is a listsuch that

∑M = 5n − k and mi ∈ {3, 4, 5} for i = 1, . . . , t, then there is a decomposition Jn →

M,k. Furthermore, for n � 13 any cycle in this decomposition having a vertex in {0, . . . , 5}has no vertex in {n, . . . , n + 5}.

Proof. We have verified using computer search and concatenation that the result holdsfor 6 � n � 21. So, let n � 22 and suppose by induction that the result holds for each positiveinteger n′ in the range 6 � n′ < n. By Lemma 47, we can assume that k � n − 1. The followingdecompositions exist by Lemma 46.

J1 −→ 51, J3 −→ 3143, J4 −→ 45, J6 −→ 310.

Case 1. Suppose that k � n − 6. Then it is routine to check, using∑

M = 5n − k � 4n + 6 �94, that M = (X,Y ), where Jy → Y is one of the decompositions above and X is some non-empty list. We can obtain a decomposition Jn → M,k by concatenating a decompositionJn−y → X, k (which exists by our inductive hypothesis, since k � n − 6 � n − y) with adecomposition Jy → Y . Since n � 22 it is clear that any 3-, 4- or 5-cycle in this decompositionhaving a vertex in {0, . . . , 5} has no vertex in {n, . . . , n + 5}, and by our inductive hypothesisthe same holds for the k-cycle.

Case 2. Suppose that n − 5 � k � n − 1. In a similar manner to Case 1, we can obtain therequired decomposition Jn → M,k if M = (X, 5) for some list X, if k ∈ {n − 5, n − 4, n − 3}and M = (X, 3, 43) for some list X, and if k ∈ {n − 5, n − 4} and M = (X, 45) for some list Y .So, we may assume that none of these hold.

Given this, using∑

M = 5n − k � 4n + 1 � 89, it is routine to check that the requireddecomposition Jn → M,k can be obtained using one of the concatenations given in the tablebelow (note that, since ν5(M) = 0, in each case we can deduce the given value of ν3(M) (mod 4)from

∑M = 5n − k). The decompositions in the third column exist by Lemma 47, and we have


verified the existence of the decompositions listed in the last column.

k ν3(M) (mod 4) First decomposition Second decomposition

n − 5 3 Jn−10 → (M − (315)), (n − 10)∗ J+10 → 315, 5+

n − 4 0 Jn−11 → (M − (316)), (n − 11)∗ J+11 → 316, 7+

n − 3 1 Jn−9 → (M − (313)), (n − 9)∗ J+9 → 313, 6+

n − 2 2 Jn−7 → (M − (310)), (n − 7)∗ J+7 → 310, 5+

Jn−5 → (M − (32, 44)), (n − 5)∗ J+5 → 32, 44, 3+

n − 1 3 Jn−8 → (M − (311)), (n − 8)∗ J+8 → 311, 7+

Jn−4 → (M − (33, 42)), (n − 4)∗ J+4 → 33, 42, 3+

Since n � 22 it is clear that any 3-, 4- or 5-cycle in this decomposition having a vertex in{0, . . . , 5} has no vertex in {n, . . . , n + 5}, and by the definition of the decompositions given inthe third column the k-cycle has no vertex in {0, . . . , 5}.

Proof of Lemma 7. We give the proof for S = {1, 2, 3, 4, 6} only. For each other connectionset S the proof proceeds along similar lines. As noted above, for n � 13 we can obtain an (M)-decomposition of 〈{1, 2, 3, 4, 6}〉n from an (M)-decomposition of Jn, provided that for eachi ∈ {0, . . . , 5}, no cycle contains both vertex i and vertex i + n. Thus, for S = {1, 2, 3, 4, 6},Lemma 7 follows by Lemma 46 for k ∈ {3, 4, 5} (since n � 13 it can be seen that no cycle inthis decomposition contains both vertex i and vertex i + n for i ∈ {0, . . . , 5}) and by Lemma 48for 6 � k � n.


In this section, we prove Lemma 9, which we restate here for convenience.

Lemma (Lemma 9). If n � 7 and M = (m1, . . . ,mt, k, n) is any list satisfying mi ∈ {3, 4, 5}for i = 1, . . . , t, 3 � k � n, and

∑M = 3n, then there is an (M)-decomposition of 〈{1, 2, 3}〉n.

The proof of Lemma 9 proceeds along similar lines to the proof of Lemma 7. We make useof the graphs J

{1,2,3}n defined in the proof of Lemma 7, which in this subsection we denote by

just Jn. Recall that J{1,2,3}n is the graph with vertex set {0, . . . , n + 2} and edge set

{{i, i + 1}, {i + 1, i + 3}, {i, i + 3} : i = 0, . . . , n − 1}.

We first construct decompositions of graphs which are related to the graphs Jn, thenconcatenate these decompositions to produce decompositions of the graphs Jn, and finallyidentify pairs of vertices to produce the required decompositions of 〈{1, 2, 3}〉n.

For n � 1, the graph obtained from Jn by adding the edges {n, n + 1} and {n + 1, n + 2}will be denoted Ln. Let M = (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t. Adecomposition {G1, . . . , Gt, A,B} of Ln such that

(i) Gi is an mi-cycle for i = 1, . . . , t;(ii) A is a path of length k from n to n + 1; and(iii) B is a path of length n − 1 from n + 1 to n + 2 such that 0, 1, 2 /∈ V (B);

will be denoted Ln → M,k+, (n − 1)H .


In Lemmas 49 and 50, we form new decompositions of graphs Ln by concatenatingdecompositions of Ln−y with decompositions of graphs Py which we will now define. Fory ∈ {3, 4, 5, 6}, the graph obtained from Jy by deleting the edges in {{0, 1}, {1, 2}} and addingthe edges in {{y, y + 1}, {y + 1, y + 2}} will be denoted Py. Let M = (m1, . . . ,mt) be a list ofintegers with mi � 3 for i = 1, . . . , t. A decomposition {G1, . . . , Gt, A1, A2, B1, B2} of Py suchthat

(i) Gi is an mi-cycle for i = 1, 2, . . . , t;(ii) A1 and A2 are vertex-disjoint paths with endpoints 0, 1, y and y + 1, such that A1 has

endpoints 0 and y or 0 and y + 1;(iii) |E(A1)| + |E(A2)| = k′ and 2 /∈ V (A1) ∪ V (A2);(iv) B1 and B2 are vertex-disjoint paths with endpoints 1, 2, y + 1 and y + 2, such that B1

has endpoints 1 and y + 1 or 1 and y + 2; and(v) |E(B1)| + |E(B2)| = y, and 0 /∈ V (B1) ∪ V (B2);

will be denoted Py → M,k′+, yH .For y ∈ {3, 4, 5, 6} and n > y, the graph Ln is the union of the graph Ln−y and the graph

obtained from Py by applying the permutation x �→ x + (n − y). It follows that if there is adecomposition Ln−y → M,k+, (n − y − 1)H and a decomposition Py → M ′, k′+, yH , then thereis a decomposition Ln → M,M ′, (k + k′)+, (n − 1)H .

Lemma 49. If n � 2 is an integer and M = (m1, . . . ,mt) is a list such that∑

M = n + 1,M /∈ {(3i) : i is even} and mi ∈ {3, 4, 5} for i = 1, . . . , t, then there is a decomposition Ln →M, (n + 2)+, (n − 1)H .

Proof. We have verified that the following decompositions exist, thus verifying the lemmafor n ∈ {2, 3, 4}.

L2 −→ 3, 4+, 1H L3 −→ 4, 5+, 2H L4 −→ 5, 6+, 3H

P4 −→ 4, 4+, 4H P5 −→ 5, 5+, 5H P6 −→ 32, 6+, 6H

So let n � 5 and assume by induction that the result holds for each integer n′ in the range2 � n′ < n. It is routine to check that for n � 5, if M satisfies the hypotheses of the lemma, thenM can be written as M = (X,Y ) where n − y � 2, X /∈ {(3i) : i is even}, and Py → Y, y+, yH

is one of the decompositions above. We can obtain the required decomposition Ln → M, (n +2)+, (n − 1)H by concatenating a decomposition Ln−y → X, (n − y + 2)+, (n − y − 1)H (whichexists by our inductive hypothesis) with a decomposition Py → Y, y+, yH .

Lemma 50. If n and k are positive integers with (4n + 12)/5 � k � n + 2, and M =(m1, . . . ,mt) is a list such that

∑M = 2n − k + 3, M /∈ {(3i) : i is even} and mi ∈ {3, 4, 5}

for i = 1, . . . , t, then there is a decomposition Ln → M,k+, (n − 1)H .

Proof. The proof will be by induction on j = n − k + 2. For a given n we need to prove theresult for each integer j in the range 0 � j � (n − 2)/5. The case j = 0 is covered in Lemma 49,so assume 1 � j � (n − 2)/5 and that the result holds for each integer j′ in the range 0 � j′ < j.Note that, since (4n + 12)/5 � k and j � 1, we have n � 7. We have verified that the followingdecompositions exist:

P3 −→ 4, 2+, 3H , P4 −→ 5, 3+, 4H , P5 −→ 32, 4+, 5H .

It is routine to check that for j � (n − 2)/5, if M satisfies the hypotheses of the lemma, thenM can be written as M = (X,Y ), where X ∈ {(3i) : i is even} and Py → Y, (y − 1)+, yH is one


of the decompositions listed above. A decomposition Ln−y → X, (k − y + 1)+, (n − y − 1)H willexist by our inductive hypothesis provided that

4(n − y) + 125

� k − y + 1 � n − y + 2

and it is routine to check that this holds using (4n + 12)/5 � k, j � 1 and y ∈ {3, 4, 5}.Thus, the required decomposition Ln → M,k+, (n − 1)H can be obtained by concatenatinga decomposition Ln−y → X, (k − y + 1)+, (n − y − 1)H with a decomposition Py → Y, (y −1)+, yH .

Let (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t. A decomposition{G1, . . . , Gt,H} of Jn such that

(i) Gi is an mi-cycle for i = 1, . . . , t; and(ii) H is an n-cycle such that 0, 1, 2 /∈ V (H) and {n, n + 2} ∈ E(H);

will be denoted Jn → m1, . . . ,mt, nH .

In Lemma 51, we will form decompositions of graphs Jn by concatenating decompositions ofgraphs Ln−y obtained from Lemma 50 with decompositions of graphs Qy which we will nowdefine. For each y ∈ {4, 5, 6}, the graph obtained from Jy by deleting the edges {0, 1} and {1, 2}will be denoted by Qy. Let M = (m1, . . . ,mt) be a list of integers with mi � 3 for i = 1, . . . , t.A decomposition {G1, . . . , Gt, A,B} of Qy such that

(i) Gi is an mi-cycle for i = 1, . . . , t;(ii) A is a path of length k′ from 0 to 1 such that {2, y, y + 1, y + 2} /∈ V (A); and(iii) B is a path of length y + 1 from 1 to 2 such that 0 ∈ V (B) and {y, y + 2} ∈ E(B);

will be denoted Qy → M,k′+, (y + 1)H .For y ∈ {4, 5, 6} and n > y, the graph Jn is the union of the graph Ln−y and the graph

obtained from Qy by applying the permutation x �→ x + (n − y). It follows that if there isa decomposition Ln−y → M,k+, (n − y − 1)H and a decomposition Qy → M ′, k′+, (y + 1)H ,then there is a decomposition Jn → M,M ′, k + k′, nH . Note that, for y ∈ {4, 5, 6} and n − y �3, no cycle of this decomposition contains both vertex i and vertex i + n for i ∈ {0, 1, 2}.

Lemma 51. If n and k are integers with n � 6, k � 3 and n − 5 � k � n and M =(m1, . . . ,mt) is a list such that

∑M = 2n − k, mi ∈ {3, 4, 5} for i = 1, . . . , t, then there is a

decomposition Jn → M,k, nH such that for i ∈ {0, 1, 2} no cycle of the decomposition containsboth vertex i and vertex i + n.

Proof. For 6 � n � 32 we have verified the result by computer search and concatenation.So, assume n � 33. The special case where M ∈ {(3i) : i is odd} will be dealt with separatelyin a moment.

Case 1. Suppose that M ∈ {(3i) : i is odd}. We have verified that the following decomposi-tions exist:

Q4 −→ 3, 2+, 5H , Q5 −→ 4, 3+, 6H , Q6 −→ 5, 4+, 7H .

It is routine to check for n � 33, if M satisfies the hypotheses of the lemma (and M /∈ {(3i) :i is odd}), then M can be written as M = (X, y − 1), where X /∈ {(3i) : i is even} and Qy →(y − 1), (y − 2)+, (y + 1)H is one of the decompositions listed above. Using n � 33 and y ∈{4, 5, 6}, it can be verified that a decomposition Ln−y → X, (k − y + 2)+, (n − y − 1)H existsby Lemma 50. Concatenation of this decomposition with Qy → (y − 1), (y − 2)+, (y + 1)H

yields the required decomposition Jn → M,k, nH .


Case 2. Suppose that M ∈ {(3i) : i is odd}. Let p = i−32 − (n − k). We deal separately with

the case n = k and the case n ∈ {k + 1, k + 2, k + 3, k + 4, k + 5}.Case 2a. Suppose that n = k. Note that since n � 33 and

∑M = 3i = 2n − k, we have p � 4

when n = k. The set of 3-cycles in the decomposition is the union of the following two sets:

{(0, 1, 3), (2, 4, 5), (n − 3, n − 2, n − 1)},{(6j + 6, 6j + 7, 6j + 8), (6j + 9, 6j + 10, 6j + 11) : j ∈ {0, . . . , p − 1}}.

The edge set of one n-cycle is E1 ∪ E2 ∪ E3, where

E1 = {{5, 3}, {3, 4}, {4, 6}},E2 = {{n − 4, n − 2}, {n − 2, n + 1}, {n + 1, n − 1}, {n − 1, n + 2}, {n + 2, n}, {n, n − 3}},E3 = {{6j + 6, 6j + 9}, {6j + 9, 6j + 8}, {6j + 8, 6j + 11},

{6j + 5, 6j + 7}, {6j + 7, 6j + 10}, {6j + 10, 6j + 12} : j ∈ {0, . . . , p − 1}}.Note that this n-cycle contains the edge {n, n + 2} and does not contain any of the vertices in{0, 1, 2}. The remaining edges form the edge set of the other n-cycle (here n = k).

Case 2b. Suppose that n ∈ {k + 1, k + 2, k + 3, k + 4, k + 5}. Since n � 33, it is routine toverify that for any integers n and k and list M ∈ {(3i) : i is odd} which satisfy the conditionsof the lemma we have p � 1, except in the case where M = (313) and n = k + 5. In this specialcase, we have (n, k) = (34, 29) and we have constructed the decomposition required in this caseexplicitly. Thus, we can assume p � 1. Let l = 5(n − k). The set of 3-cycles in the decompositionis the union of the following three sets:

{(0, 1, 3), (2, 4, 5), (n − 3, n − 2, n − 1)},{(5j + 6, 5j + 7, 5j + 8), (5j + 9, 5j + 10, 5j + 11) : j ∈ {0, . . . , (n − k) − 1}},

{(6j + l + 6, 6j + l + 7, 6j + l + 8), (6j + l + 9, 6j + l + 10, 6j + l + 11) : j ∈ {0, . . . , p − 1}}.The edge set of the n-cycle is E1 ∪ E2 ∪ E3 ∪ E4, where

E1 = {{5, 3}, {3, 4}, {4, 6}},E2 = {{n − 4, n − 2}, {n − 2, n + 1}, {n + 1, n − 1}, {n − 1, n + 2}, {n + 2, n}, {n, n − 3}},E3 = {{5j + 6, 5j + 9}, {5j + 9, 5j + 8}, {5j + 8, 5j + 11},

{5j + 5, 5j + 7}, {5j + 7, 5j + 10} : j ∈ {0, . . . , (n − k) − 1}},E4 = {{6j + l + 6, 6j + l + 9}, {6j + l + 9, 6j + l + 8},

{6j + l + 8, 6j + l + 11}, {6j + l + 5, 6j + l + 7},{6j + l + 7, 6j + l + 10}, {6j + l + 10, 6j + l + 12} : j ∈ {0, . . . , p − 1}}.

Note that this n-cycle contains the edge {n, n + 2} and does not contain any of the vertices in{0, 1, 2}. The remaining edges form the edge set of the cycle of length k.

In Lemma 52, we will form decompositions of graphs Jn by concatenating decompositionsof Jn−y with decompositions of graphs Ry which we will now define. For y ∈ {5, 6}, the graphobtained from Jy by adding the edge {0, 2} will be denoted by Ry. Let M = (m1, . . . ,mt) be alist of integers with mi � 3 for i = 1, . . . , t. A decomposition {G1, . . . , Gt, A} of Ry such that

(i) Gi is an mi-cycle for i = 1, . . . , t; and(ii) A is a path of length y + 1 from 0 to 2 such that 1 ∈ V (A) and {y, y + 2} ∈ E(A);

will be denoted Ry → M,yH .For y ∈ {5, 6} and n > y, the graph Jn is the union of the graph obtained from Jn−y

by removing the edge {n − y, n − y + 2} and the graph obtained from Ry by applying thepermutation x �→ x + (n − y). It follows that if there is a decomposition Jn−y → M,k, (n − y)H


and a decomposition Ry → M ′, yH , then there is a decomposition Jn → M,M ′, k, nH . In thisconstruction, the edge {n − y, n − y + 2} in the (n − y)-cycle of the decomposition of Jn−y isreplaced with the path from the decomposition of Ry to form the n-cycle in the decompositionof Jn. Note that, for y ∈ {5, 6} and n − y � 3, no cycle of the decomposition contains bothvertex i and vertex i + n for i ∈ {0, 1, 2}.

Lemma 52. Let n and k be integers with 6 � k � n. If M = (m1, . . . ,mt) is a list such that∑M = 2n − k and mi ∈ {3, 4, 5} for i = 1, . . . , t, then there is a decomposition Jn → M,k, nH

such that for i ∈ {0, 1, 2} no cycle of the decomposition contains both vertex i and vertex i + n.

Proof. If k � n − 5, then the result follows by Lemma 51, which means the result holdsfor n � 11. We can therefore assume that k � n − 6, n � 12 and, by induction, that the resultholds for each integer n′ in the range 6 � n′ < n.

We have verified that the following decompositions exist:

R5 −→ 52, 5H , R6 −→ 3, 4, 5, 6H , R6 −→ 43, 6H , R6 −→ 34, 6H .

It is routine to check that if M satisfies the conditions of the lemma, then M can bewritten as M = (X,Y ), where Ry → Y, yH is one of the decompositions above. The requireddecomposition can be obtained by concatenating a decomposition Jn−y → X, k, (n − y)H

(which exists by our inductive hypothesis since k � n − 6 � n − y) with a decompositionRy → Y, yH .

Proof of Lemma 9. If k ∈ {3, 4, 5}, then the result follows by Lemma 7. So, we canassume k � 6. Since n � 7, we can obtain an (M)-decomposition of 〈{1, 2, 3}〉n from an (M)-decomposition of Jn by identifying vertex i with vertex i + n for each i ∈ {0, 1, 2}, providedthat for each i ∈ {0, 1, 2}, no cycle of our decomposition contains both vertex i and vertexi + n. Thus, Lemma 9 follows immediately from Lemma 52.

7. Decompositions of Kn − 〈S〉nThe purpose of this section is to prove Lemmas 11 and 12, and these proofs are givenin Subsections 7.2 and 7.3, respectively. In Subsection 7.1, we present results on Hamiltondecompositions of circulant graphs that we will require.

To prove Lemma 11, we require a (3tn, 4qn, nh)-decomposition of 〈S〉n, where S ={1, . . . , �n/2�} \ S, for almost all n, t, q and h satisfying h � 2, n � 2max(S) + 1 and 3t + 4q +h = �(n − 1)/2� − |S|. To construct this, S will be partitioned into three subsets S1, S2 and S3

such that there is a (3tn)-decomposition of 〈S1〉n, a (4qn)-decomposition of 〈S2〉n, and an (nh)-decomposition of 〈S3〉n. Our (3tn)-decompositions of 〈S1〉n are constructed by partitioningS1 into modulo n difference triples, our (4qn)-decompositions of 〈S2〉n are constructed bypartitioning S2 into modulo n difference quadruples, and our (nh)-decompositions of 〈S3〉n areconstructed by partitioning S3 into sets of size at most 3 to yield connected circulant graphsof degree at most 6 that are known to have Hamilton decompositions. Lemma 12 is proved inan analogous manner.

7.1. Decompositions of circulant graphs into Hamilton cycles

Theorems 53–55 address the open problem of whether every connected Cayley graph on a finiteabelian group has a Hamilton decomposition [4]. Note that 〈S〉n is connected if and only ifgcd(S ∪ {n}) = 1.


Theorem 53 [13]. Every connected 4-regular Cayley graph on a finite abelian group hasa decomposition into two Hamilton cycles.

The following theorem is an easy corollary of Theorem 53.

Theorem 54. Every connected 5-regular Cayley graph on a finite abelian group has adecomposition into two Hamilton cycles and a perfect matching.

Proof. Let the graph be X = Cay(Γ, S). Since each vertex of X has odd degree, S containsan element s of order 2 in Γ. Let F be the perfect matching of X generated by s. IfCay(Γ, S \ {s}) is connected then, as it is also 4-regular, the result follows immediately fromTheorem 53. On the other hand, if Cay(Γ, S \ {s}) is not connected, then it consists of twoisomorphic connected components, with x �→ sx being an isomorphism. These components are4-regular and so by Theorem 53, each can be decomposed into two Hamilton cycles. Moreover,since x �→ sx is an isomorphism, there exists a Hamilton decomposition {H1,H

′1} of the first

and a Hamilton decomposition {H2,H′2} of the second such that there is a pair of vertex-disjoint

4-cycles (x1, y1, y2, x2) and (x′1, y

′1, y

′2, x

′2) in X with x1y1 ∈ E(H1), x′

1y′1 ∈ E(H ′

1), x2y2 ∈E(H2), x′

2y′2 ∈ E(H ′

2), and x1x2, y1y2, x′1x

′2, y

′1y

′2 ∈ E(F ). It follows that if we let G be the

graph with edge set

(E(H1) ∪ E(H2) ∪ {x1x2, y1y2}) \ {x1y1, x2y2}

and let G′ be the graph with edge set

(E(H ′1) ∪ E(H ′

2) ∪ {x′1x

′2, y

′1y

′2}) \ {x′

1y′1, x

′2y

′2},

then G and G′ are edge-disjoint Hamilton cycles in X. This proves the result.

Theorem 55 [31]. Every 6-regular Cayley graph on a group which is generated by anelement of the connection set has a decomposition into three Hamilton cycles.

This theorem implies that, for distinct a, b, c ∈ {1, . . . , �(n − 1)/2�}, the graph 〈{a, b, c}〉nhas a decomposition into three Hamilton cycles if gcd(x, n) = 1 for some x ∈ {a, b, c}.

In the next two lemmas we give results similar to that of Lemma 10, but for the casewhere the connection set is of the form {x − 1} ∪ {x + 1, . . . , �n/2�} rather than {x, . . . , �n/2�}.Lemma 56 deals with the case n is odd, and Lemma 57 deals with the case n is even.

Lemma 56. If n is odd and 1 � h � (n − 3)/2, then there is an (nh)-decomposition of〈{(n − 1)/2 − h} ∪ {(n − 1)/2 − h + 2, . . . , (n − 1)/2}〉n; except when h = 1 and n ≡ 3 (mod 6)in which case the graph is not connected.

Proof. If h = 1, then the graph is 〈(n − 3)/2〉n. If n ≡ 1, 5 (mod 6), then gcd((n − 3)/2, n) =1 and 〈(n − 3)/2〉n is an n-cycle. If n ≡ 3 (mod 6), then gcd((n − 3)/2, n) = 3 and 〈(n − 3)/2〉nis not connected. Thus the result holds for h = 1. In the remainder of the proof we assume h � 2.

We first decompose 〈{(n − 1)/2 − h} ∪ {(n − 1)/2 − h + 2, . . . , (n − 1)/2}〉n into circulantgraphs by partitioning the connection set, and then decompose the resulting circulant graphsinto n-cycles using Theorems 53 and 55.

If h is even, then we partition the connection set into pairs by pairing (n − 1)/2 − h with(n − 1)/2 and partitioning {(n − 1)/2 − h + 2, . . . , (n − 3)/2} into consecutive pairs (if h = 2,


then our partition is just {{(n − 5)/2, (n − 1)/2}}). Each of the resulting circulant graphs is4-regular and connected and thus can be decomposed into two n-cycles by Theorem 53. Ifh is odd, then we partition the connection set into the triple {(n − 1)/2 − h, (n − 1)/2 − h +2, (n − 1)/2} and consecutive pairs from {(n − 1)/2 − h + 3, . . . , (n − 3)/2} (if h = 3, then ourpartition is just {{(n − 7)/2, (n − 3)/2, (n − 1)/2}}). Since gcd((n − 1)/2, n) = 1, the graph〈{(n − 1)/2 − h, (n − 1)/2 − h + 2, (n − 1)/2}〉n can be decomposed into three n-cycles byTheorem 55. Any other resulting circulant graphs are 4-regular and connected and thus caneach be decomposed into two n-cycles by Theorem 53.

Lemma 57. If n is even and 1 � h � (n − 4)/2, then there is an (nh)-decomposition of〈{n/2 − h − 1} ∪ {n/2 − h + 1, . . . , n/2}〉n; except when h = 1 and n ≡ 0 (mod 4) in which casethe graph is not connected.

Proof. If h = 1, then the graph is 〈{(n − 4)/2, n/2}〉n. If n ≡ 2 (mod 4), thengcd((n − 4)/2, n) = 1, 〈(n − 4)/2〉n is an n-cycle, and 〈{n/2}〉n is a perfect matching. Ifn ≡ 0 (mod 4), then gcd((n − 4)/2, n

2 , n) = 2 and 〈{(n − 4)/2, n/2}〉n is not connected. Thusthe result holds for h = 1. In the remainder of the proof, we assume h � 2.

We first decompose 〈{n/2 − h − 1} ∪ {n/2 − h + 1, . . . , n/2}〉n into circulant graphs bypartitioning the connection set, and then decompose the resulting circulant graphs into n-cyclesusing Theorems 53–55.

If n ≡ 0 (mod 4) and h is even, then we partition the connection set into pairs andthe singleton {n/2} by pairing n/2 − h − 1 with (n − 2)/2 and partitioning {n/2 − h +1, . . . , (n − 4)/2} into pairs of consecutive integers (if h = 2, then our partition is just{{n/2}, {(n − 6)/2, (n − 2)/2}}). The graph 〈{n/2}〉n is a perfect matching. The otherresulting circulant graphs are 4-regular and connected and thus can each be decomposed intotwo n-cycles by Theorem 53 (note that gcd((n − 2)/2, n) = 1).

If n ≡ 0 (mod 4) and h is odd, then we partition the connection set into pairs, thetriple {n/2 − h − 1, n/2 − h + 1, (n − 2)/2} and the singleton {n/2} by partitioning {n/2 −h + 2, . . . , (n − 4)/2} into pairs of consecutive integers (if h = 3, then our partition is just{{n/2}, {(n − 8)/2, (n − 4)/2, (n − 2)/2}}). The graph 〈{n/2}〉n is a perfect matching and,since gcd((n − 2)/2, n) = 1, the graph 〈{n/2 − h − 1, n/2 − h + 1, (n − 2)/2}〉n can be decom-posed into three n-cycles using Theorem 55. Any other resulting circulant graphs are 4-regularand connected and thus can each be decomposed into two n-cycles by Theorem 53.

If n ≡ 2 (mod 4), then we partition the connection set into pairs, the triple {n/2 − h −1, (n − 2)/2, n/2} and, when h is odd, the singleton {(n − 4)/2} by partitioning {n/2 −h + 1, . . . , (n − 4)/2} into pairs of consecutive integers (when h is even) or into pairs ofconsecutive integers and the singleton {(n − 4)/2} (when h is odd). (Our partition isjust {{(n − 6)/2, (n − 2)/2, n/2}} if h = 2, and just {{(n − 8)/2, (n − 2)/2, n/2}, {(n − 4)/2}}if h = 3.) Since gcd((n − 2)/2, n/2) = 1, the graph 〈{n/2 − h − 1, (n − 2)/2, n/2}〉n can bedecomposed into two n-cycles and a perfect matching using Theorem 54. When h is odd,〈{(n − 4)/2}〉n is an n-cycle (note that gcd((n − 4)/2, n) = 1). Any other resulting circulantgraphs are 4-regular and connected and thus can each be decomposed into two n-cycles byTheorem 53.


Before we prove Lemma 11, we require three preliminary lemmas which establish the existenceof various (4qn, nh)-decompositions of circulant graphs.


Lemma 58. If S ⊆ {1, . . . , �(n − 1)/2�} such that

(i) S = {x + 1, . . . , x + 4q} for some x;(ii) S = {x} ∪ {x + 2, . . . , x + 4q − 1} ∪ {x + 4q + 1} for some x; or(iii) S = {(n − 1)/2 − 4q} ∪ {(n − 1)/2 − 4q + 2, . . . , (n − 1)/2} where n is odd;

then there is a (4qn)-decomposition of 〈S〉n.

Proof. It is sufficient to partition S into q modulo n difference quadruples. If S = {x +1, . . . , x + 4q}, then we partition S into q sets of the form {y, y + 1, y + 2, y + 3}, each of whichis a difference quadruple. If S = {x} ∪ {x + 2, . . . , x + 4q − 1} ∪ {x + 4q + 1}, then we partitionS into q sets of the form {y, y + 2, y + 3, y + 5}, each of which is a difference quadruple. IfS = {(n − 1)/2 − 4q} ∪ {(n − 1)/2 − 4q + 2, . . . , (n − 1)/2} and n is odd, then we partition Sinto q − 1 sets of the form {y, y + 2, y + 3, y + 5}, each of which is a difference quadruple, andthe set {(n − 9)/2, (n − 5)/2, (n − 3)/2, (n − 1)/2}, which is a modulo n difference quadruple(note that (n − 5)/2 + (n − 3)/2 + (n − 1)/2 − (n − 9)/2 = n).

Lemma 59. If h, q and n are non-negative integers with 1 � 4q + h � �(n − 1)/2�, thenthere is a (4qn, nh)-decomposition of 〈{�(n − 1)/2� − h − 4q + 1, . . . , �n/2�}〉n.

Proof. If h = 0, then the result follows immediately by Lemma 58, and if q = 0, then theresult follows immediately by Lemma 10. For q, h � 1 we partition the connection set intothe set {�(n − 1)/2� − h − 4q + 1, . . . , �(n − 1)/2� − h} and the set {�(n − 1)/2� − h + 1, . . . ,�n/2�}. Then 〈{�(n − 1)/2� − h − 4q + 1, . . . , �(n − 1)/2� − h}〉n has a (4qn)-decompositionby Lemma 58, and 〈{�(n − 1)/2� − h + 1, . . . , �n/2�}〉n has an (nh)-decomposition byLemma 10.

Lemma 60. If h, q and n are non-negative integers with 1 � 4q + h � �(n − 3)/2� suchthat n is odd when h = 0 and n ≡ 1, 2, 5, 6, 7, 10, 11 (mod 12) when h = 1, then there is a(4qn, nh)-decomposition of 〈{�(n − 1)/2� − h − 4q} ∪ {�(n − 1)/2� − h − 4q + 2, . . . , �n/2�}〉n.

Proof. If h = 0, then the result follows immediately by Lemma 58. If q = 0, then theresult follows immediately by Lemma 56 (n odd) or Lemma 57 (n even). For h, q � 1,we partition the connection set into the set {�(n − 1)/2� − h − 4q} ∪ {�(n − 1)/2� −h − 4q + 2, . . . , �(n − 1)/2� − h − 1} ∪ {�(n − 1)/2� − h + 1} and the set {�(n − 1)/2� − h} ∪{�(n − 1)/2� − h + 2, . . . , �n/2�}. Then 〈{�(n− 1)/2� − h − 4q} ∪ {�(n− 1)/2� − h− 4q +2, . . . , �(n− 1)/2� − h − 1} ∪ {�(n − 1)/2� − h + 1}〉n has a (4qn)-decomposition by Lemma 58,and 〈{�(n − 1)/2� − h} ∪ {�(n − 1)/2� − h + 2, . . . , �n/2�}〉n has an (nh)-decomposition byLemma 56 (n odd) or Lemma 57 (n even).

Proof of Lemma 11. We give the proof of Lemma 11 for the case S = {1, 2, 3, 4, 6} only.The cases

S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}}

are similar.The conditions h � 2 and 3t + 4q + h = �(n − 1)/2� − 5 imply n � 6t + 15. If t = 0, then

the result follows immediately by Lemma 60. We deal separately with the three cases t ∈{1, 2, 3, 4, 5, 6}, t ∈ {7, 8, 9, 10}, and t � 11.


Case 1. Suppose that t ∈ {1, 2, 3, 4, 5, 6}. The cases 6t + 15 � n � 6t + 18 and the cases

(n, t) ∈ {(38, 3), (39, 3), (40, 3), (44, 4), (45, 4)}

are dealt with first. Since h � 2, it follows from 3t + 4q + h = �(n − 1)/2� − 5 that in each ofthese cases we have q = 0. Thus, the value of h is uniquely determined by the values of n and t.The required decompositions are obtained by partitioning {5} ∪ {7, . . . , �n/2�} into t modulo ndifference triples and a collection of connection sets for circulant graphs such that the circulantgraphs can be decomposed into Hamilton cycles (or Hamilton cycles and a perfect matching)using the results in Section 7.1. Suitable partitions are given in the following tables.

t = 1:

n Modulo n Connection setsdifference triples

21 {5, 7, 9} {8, 10}22 {5, 7, 10} {8, 9}, {11}23 {5, 8, 10} {7, 9}, {11}24 {5, 9, 10} {7, 8}, {11}, {12}

t = 2:


27 {5, 7, 12}, {8, 9, 10} {11, 13}28 {5, 7, 12}, {8, 9, 11} {10, 13}, {14}29 {5, 7, 12}, {8, 10, 11} {9}, {13, 14}30 {5, 9, 14}, {8, 10, 12} {7}, {11, 13}, {15}

t = 3:


33 {5, 8, 13}, {7, 9, 16}, {10, 11, 12} {14, 15}34 {5, 10, 15}, {7, 9, 16}, {8, 12, 14} {11, 13}, {17}35 {5, 8, 13}, {7, 9, 16}, {10, 11, 14} {12, 15}, {17}36 {5, 8, 13}, {7, 9, 16}, {10, 12, 14} {11, 15}, {17}, {18}38 {5, 12, 17}, {7, 9, 16}, {8, 10, 18} {11, 13}, {14, 15}, {19}39 {5, 12, 17}, {7, 9, 16}, {8, 10, 18} {11, 13}, {14, 15}, {19}40 {5, 12, 17}, {7, 9, 16}, {8, 10, 18} {11, 13}, {14, 15}, {19}, {20}

t = 4:


39 {5, 11, 16}, {7, 8, 15}, {9, 10, 19}, {12, 13, 14} {17, 18}40 {5, 8, 13}, {7, 9, 16}, {10, 12, 18}, {11, 14, 15} {17, 19}, {20}41 {5, 14, 19}, {7, 8, 15}, {9, 11, 20}, {12, 13, 16} {10, 17}, {18}42 {5, 10, 15}, {7, 11, 18}, {8, 9, 17}, {12, 14, 16} {13}, {19, 20}, {21}44 {5, 11, 16}, {7, 13, 20}, {8, 10, 18}, {9, 12, 21} {14, 15}, {17, 19}, {22}45 {5, 11, 16}, {7, 13, 20}, {8, 10, 18}, {9, 12, 21} {14, 15}, {17, 19}, {22}


t = 5:


45 {5, 17, 22}, {7, 13, 20}, {8, 10, 18}, {9, 12, 21}, {14, 15, 16} {11, 19}46 {5, 17, 22}, {7, 13, 20}, {8, 10, 18}, {9, 12, 21}, {11, 16, 19} {14, 15}, {23}47 {5, 18, 23}, {7, 13, 20}, {8, 11, 19}, {10, 12, 22}, {14, 16, 17} {9, 15}, {21}48 {5, 17, 22}, {7, 13, 20}, {8, 10, 18}, {9, 12, 21}, {11, 14, 23} {15, 16}, {19}, {24}

t = 6:n Modulo n Connection sets

difference triples51 {5, 13, 18}, {7, 15, 22}, {8, 16, 24}, {21, 25}

{9, 10, 19}, {11, 12, 23}, {14, 17, 20}52 {5, 13, 18}, {7, 15, 22}, {8, 16, 24}, {20, 25}, {26}

{9, 10, 19}, {11, 12, 23}, {14, 17, 21}53 {5, 13, 18}, {7, 12, 19}, {8, 14, 22}, {23}, {25, 26}

{9, 15, 24}, {10, 11, 21}, {16, 17, 20}54 {5, 17, 22}, {7, 8, 15}, {9, 10, 19}, {21, 23}, {25}, {27}

{11, 13, 24}, {12, 14, 26}, {16, 18, 20}

We now deal with n � 6t + 19 which implies 4q + h � 4. Define St by St = {5} ∪ {7, . . . , 3t +9} for t ∈ {1, 2, 5, 6}, and St = {5} ∪ {7, . . . , 3t + 8} ∪ {3t + 10} when t ∈ {3, 4}. The followingtable gives a partition πt of St into difference triples and a difference quadruple Qt such thatQt can be partitioned into two pairs of relatively prime integers.

t πt

1 {{5, 7, 12}, {8, 9, 10, 11}}2 {{5, 9, 14}, {7, 8, 15}, {10, 11, 12, 13}}3 {{5, 9, 14}, {7, 10, 17}, {8, 11, 19}, {12, 13, 15, 16}}4 {{5, 9, 14}, {7, 13, 20}, {8, 11, 19}, {10, 12, 22}, {15, 16, 17, 18}}5 {{5, 14, 19}, {7, 13, 20}, {8, 10, 18}, {9, 15, 24}, {11, 12, 23}, {16, 17, 21, 22}}6 {{5, 15, 20}, {7, 16, 23}, {8, 14, 22}, {9, 12, 21}, {10, 17, 27}, {11, 13, 24},

{18, 19, 25, 26}}

Thus, 〈Qt〉n can be decomposed into two connected 4-regular Cayley graphs, which inturn can be decomposed into Hamilton cycles using Theorem 53. It follows that there isboth a (3tn, 4n)-decomposition and a (3tn, n4)-decomposition of 〈St〉n. If q = 0, then we usethe (3tn, n4)-decomposition of 〈St〉n and if q � 1, then we use the (3tn, 4n)-decompositionof 〈St〉n. This leaves us needing an (nh−4)-decomposition of Kn − 〈{1, 2, 3, 4, 6} ∪ St〉n whenq = 0, and a (4(q−1)n, nh)-decomposition of Kn − 〈{1, 2, 3, 4, 6} ∪ St〉n when q � 1. Note thatKn − 〈{1, 2, 3, 4, 6} ∪ St〉n is isomorphic to

(i) 〈{3t + 10, . . . , �n/2�}〉n when t ∈ {1, 2, 5, 6}; and(ii) 〈{3t + 9} ∪ {3t + 11, . . . , �n/2�}〉n when t ∈ {3, 4}.When t ∈ {1, 2, 5, 6} the required decomposition exists by Lemma 59. When t ∈ {3, 4} and

the required number of Hamilton cycles (that is, h − 4 when q = 0 and h when q � 1) is atleast 2, the required decomposition exists by Lemma 60. So, we need to consider only the caseswhere q = 0, h ∈ {4, 5} and t ∈ {3, 4}.

Since 3t + 4q + h = �(n − 1)/2� − 5, and since we have already dealt with the cases where(n, t) ∈ {(38, 3), (39, 3), (40, 3), (44, 4), (45, 4)}, this leaves us with only the three cases where(n, t, h) ∈ {(37, 3, 4), (43, 4, 4), (46, 4, 5)}. In the cases (n, t, h) ∈ {(37, 3, 4), (43, 4, 4)}, we have


that h − 4 (the required number of Hamilton cycles) is 0 and n is odd, and in the case (n, t, h) =(46, 4, 5) we have that h − 4 (the required number of Hamilton cycles) is 1 and n ≡ 10 (mod 12).So, in all these cases the required decompositions exist by Lemma 60.

Case 2. Suppose that t ∈ {7, 8, 9, 10}. Redefine St by St = {5} ∪ {7, . . . , 3t + 7}. Thefollowing table gives a partition of St into difference triples and a set Rt consisting ofa pair of relatively prime integers. Thus, 〈Rt〉n is a connected 4-regular Cayley graph,and so can be decomposed into two Hamilton cycles using Theorem 53. Thus, we have a(3tn, n2)-decomposition of 〈St〉n.

t Difference triples Rt

7 {5, 17, 22}, {7, 16, 23}, {8, 20, 28}, {9, 18, 27}, {10, 14, 24}, {11, 15, 26}, {19, 21}{12, 13, 25}

8 {5, 17, 22}, {7, 19, 26}, {8, 16, 24}, {9, 20, 29}, {10, 21, 31}, {11, 14, 25}, {23, 27}{12, 18, 30}, {13, 15, 28}

9 {5, 19, 24}, {7, 20, 27}, {8, 21, 29}, {9, 22, 31}, {10, 23, 33}, {11, 15, 26}, {25, 34}{12, 16, 28}, {13, 17, 30}, {14, 18, 32}

10 {5, 21, 26}, {7, 22, 29}, {8, 23, 31}, {9, 24, 33}, {10, 25, 35}, {11, 17, 28}, {32, 37}{12, 18, 30}, {13, 14, 27}, {15, 19, 34}, {16, 20, 36}

Thus, we only require a (4qn, nh−2)-decomposition of Kn − 〈{1, 2, 3, 4, 6} ∪ St〉n. But Kn −〈{1, 2, 3, 4, 6} ∪ St〉n is isomorphic to 〈{3t + 8, . . . , �n/2�}〉n and so this decomposition existsby Lemma 59.

Case 3. Suppose that t � 11. Redefine St by St = {5} ∪ {7, . . . , 3t + 5} when t ≡ 1, 2 (mod 4),and St = {5} ∪ {7, . . . , 3t + 4} ∪ {3t + 6} when t ≡ 0, 3 (mod 4). We now obtain a (3tn)-decomposition of 〈St〉n. For 11 � t � 101, we have found such a decomposition by partitioningSt into difference triples with the aid of a computer. For t � 102, we first decompose 〈St〉n into〈{5} ∪ {7, . . . , 44}〉n = 〈S13〉n and 〈St \ S13〉n. We have already noted the existence of a (313n)-decomposition of 〈S13〉n. For t ≡ 1, 2 (mod 4) (respectively, t ≡ 0, 3 (mod 4)), we can obtain a(3(t−13)n)-decomposition of 〈St \ S13〉n by using a Langford sequence (respectively, hookedLangford sequence) of order t − 13 and defect 45, which exists since t � 102, to partitionSt \ S13 into difference triples (see [47, 48]). So we have a (3tn)-decomposition of 〈St〉n,and require a (4qn, nh)-decomposition of Kn − 〈{1, 2, 3, 4, 6} ∪ St〉n. Since Kn − 〈{1, 2, 3, 4, 6} ∪St〉n is isomorphic to

(i) 〈{3t + 6, . . . , �n/2�}〉n when t ≡ 1, 2 (mod 4); and(ii) 〈{3t + 5} ∪ {3t + 7, . . . , �n/2�}〉n when t ≡ 0, 3 (mod 4);

this decomposition exists by Lemma 59 or 60.


Proof of Lemma 12. We give the proof for the case S = {1, 2, 3, 4, 6} only. The proofs forthe cases

S ∈ {{1, 2, 3, 4}, {1, 2, 3, 4, 5, 7}, {1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7, 8}}

are similar.The conditions h � 2 and 5r + h = �(n − 1)/2� − 5 imply n � 10r + 15. If r = 0, then the

result follows immediately by Lemma 56 (n odd) or Lemma 57 (n even). Thus, we assume r � 1.Define Sr by Sr = {5} ∪ {7, . . . , 5r + 5} when r ≡ 2, 3 (mod 4), and Sr = {5} ∪ {7, . . . , 5r +

4} ∪ {5r + 6} when r ≡ 0, 1 (mod 4). We now obtain a (5rn)-decomposition 〈Sr〉n by partition-ing Sr into difference quintuples. We have constructed such a partition of S for 1 � r � 30with the aid of a computer.


For r � 31, we first partition Sr into {5} ∪ {7, . . . , 15} = S2 and Sr \ S2. We have alreadynoted that S2 can be partitioned into difference quintuples, so we only need to partition Sr \ S2

into difference quintuples. Note that Sr \ S2 = {16, . . . , 5r + 5} when r ≡ 2, 3 (mod 4), andSr \ S2 = {16, . . . , 5r + 4} ∪ {5r + 6} when r ≡ 0, 1 (mod 4).

For r ≡ 2, 3 (mod 4), we take a Langford sequence of order r − 2 and defect 15, whichexists since r � 31 (see [47, 48]), and use it to partition {15, . . . , 3r + 8} into r − 2 differencetriples. We then add 1 to each element of each of these triples to obtain a partition of{16, . . . , 3r + 9} into r − 2 triples of the form {a, b, c} where a + b = c + 1. It is easy to constructthe required partition of {16, . . . , 5r + 5} into difference quintuples from this by partitioning{3r + 10, . . . , 5r + 5} into r − 2 pairs of consecutive integers.

For r ≡ 0, 1 (mod 4), we take a hooked Langford sequence of order r − 2 and defect 15, whichexists since r � 31 (see [47, 48]), and use it to partition {15, . . . , 3r + 7} ∪ {3r + 9} into r − 2difference triples. We then add 1 to each element of each of these triples, except the element3r + 9, to obtain a partition of {16, . . . , 3r + 9} into r − 3 triples of the form {a, b, c} wherea + b = c + 1, and one triple of the form {a, b, c} where a + b = c + 2. It is easy to constructthe required partition of {16, . . . , 5r + 4} ∪ {5r + 6} into difference quintuples from this bypartitioning {3r + 10, . . . , 5r + 4} ∪ {5r + 6} into r − 3 pairs of consecutive integers, and thepair {5r + 4, 5r + 6}. The pair {5r + 4, 5r + 6} combines with the triple of the form {a, b, c},where a + b = c + 2 to form a difference quintuple.

So, we have a (5rn)-decomposition of 〈Sr〉n, and require an (nh)-decomposition of Kn −〈{1, 2, 3, 4, 6} ∪ Sr〉n. Note that Kn − 〈{1, 2, 3, 4, 6} ∪ Sr〉n is isomorphic to

(i) 〈{5r + 6, . . . , �n/2�}〉n when r ≡ 2, 3 (mod 4); and(ii) 〈{5r + 5} ∪ {5r + 7, . . . , �n/2�}〉n when r ≡ 0, 1 (mod 4).

If r ≡ 2, 3 (mod 4), then the decomposition exists by Lemma 10. If r ≡ 0, 1 (mod 4), then thedecomposition exists by Lemma 56 (n odd) or Lemma 57 (n even).

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Darryn Bryant and William PetterssonSchool of Mathematics and PhysicsThe University of QueenslandQueensland 4072Australia

db@maths·uq·edu·[email protected]

Daniel HorsleySchool of Mathematical SciencesMonash UniversityVictoria 3800Australia

danhorsley@gmail·com

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Cycle decompositions V: Complete graphs into cycles of arbitrary lengths